Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap10 Dynamics in 1D 12 Particles in space

Chapter 11 Vibrations

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11.0Introductionstudy PDF
11.1Damped vibrationsstudy PDF + solutions
11.2Forcing & resonancestudy PDF + solutions
11.3Normal modesstudy PDF + solutions

The ideal harmonic oscillator (previous chapter) oscillates with simple sinusoidal motion forever. In the real world there is friction so oscillations decay more or less quickly, there are forces that cause the oscillations, and there are multiple parts that move multiple ways. Understanding something about these realities is the topic of this chapter. Some key words are damping, resonance, frequency response and normal modes.

This chapter is a brief introduction to the topic of vibrations. We already know about the harmonic oscillator, and that is the core idea. But here we want to take account of three main ideas, with one section per idea.

1.

Damping.There is friction, so oscillatory motion decays in time. This is simple enough conceptually, but the mathematical descriptions are more complicated than for the harmonic oscillator, involving a mix of exponential and sine functions.
2.

Forcing. Things get pushed around, or ‘forced’. What is the relation between how much you push on something and how much it shakes. The key ideas here are resonance and frequency response.
3.

Normal modes. Most real machines and structures have various moving parts. How do things move with various parts? For simplicity here we stick to the cases with no friction. The key idea is that one can think of the general motions as made up of (as a superposition of) simple harmonic motions. These are called normal modes.

11.1 Damped vibrations

In the real world, macroscopic oscillators that are not pumped have motions that decay in time due to dissipation. Our goal is to understand what happens to a harmonic oscillator when we add friction. In particular we are interested in the simplest kind of friction caused by an ideal linear dashpot. What we will find is exponentially decaying solutions that may, or may not have, an oscillatory nature depending on the amount of friction.

Damping

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Figure 11.1: A dashpot. A dashpot (or damper) is shown here connecting two parts of a mechanism. The tension in the dashpot is proportional to the rate at which it lengthens. The symbol shown represents any device which resists the relative motion of its endpoints. The schematic is supposed to suggest a plunger in a cylinder. For the plunger to move, fluid must leak around the cylinder. This leakage happens for either direction of motion. Thus the damper resists relative motion in either direction; i. e., for

\dot{\ell}>0

and

\dot{\ell}<0

Dashpots are used to absorb energy. One is shown schematically in fig. 11.1. Often springs and dashpots are light in comparison to the machinery to which they are attached so their mass and weight are neglected. They are usually attached with pin joints, ball and socket joints, or other kinds of flexible connections so only forces are transmitted. Because they only have forces at their ends they are ‘two-force’ bodies so (see section 5.2) the forces at their ends are equal, opposite, and along the line of connection.

The most familiar examples are the shock absorbers of a car or the damper for screen doors.

The dashpot provides resistance to motion by sucking or pushing air or oil in and out of the cylinder through a small opening. Due to the viscosity of the air or oil, a pressure drop is created across the small opening. Ideally, this viscous resistance causes a force that is exactly proportional to the velocity, giving so-called linear damping. The same proportionality, of force and speed, is assumed to hold for both lengthening and shortening. So, the compression force (negative tension) is also proportional to the rate at which the dashpot shortens (negative lengthens).

The tension in the dashpot is usually assumed to be proportional to the rate at which it lengthens, although this approximation is not especially accurate for most dampers one can buy. In a physical dashpot nonlinearities, from the fluid flow and from friction between the piston and the cylinder, are often significant. Also, dashpots that use air as a working fluid may have compressibility that introduces extra springiness to the system.

The defining equation for an ideal linear dashpot is:

T=C\,\dot{\ell}

where C is the dashpot constant.

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Figure 11.2: The effect of varying the damping with a fixed mass and spring. In all the plots the mass is released from rest at

x=x_{0}

. In the case of under-damping, oscillations persist for a long time, forever if there is no damping. In the case of over-damping, the dashpot doesn’t relax for a long time; it stays locked up forever in the limit of

c\rightarrow\infty

. The fastest relaxation occurs for critical damping.

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Figure 11.3: A mass spring dashpot system, or damped harmonic oscillator. The free-body diagram of the mass shows the tension in the dashpot.

Damped oscillations

Adding a dashpot in parallel with the spring of a mass-spring system creates a mass-spring-dashpot system, or damped harmonic oscillator. The system is shown in fig. 11.3.

Also in fig. 11.3 is a free-body diagram of the mass. It has two forces acting on it, neglecting gravity:

	$\displaystyle F_{s}$	$\displaystyle=$	$\displaystyle kx$	is the spring force, assuming a linear spring, and
	$\displaystyle F_{d}$	$\displaystyle=$	$\displaystyle c\ dx/dt=c\dot{x}$	is the dashpot force assuming a linear dashpot.

The system is a one degree of freedom system because a single coordinate $x$ is sufficient to describe the complete motion of the system.

The equation of motion for this system is

m\ddot{x}=-F_{d}-F_{s}\quad\hbox to36.135pt{where\hfil}\ddot{x}=d^{2}x/dt^{2}.

(11.1)

Assuming a linear spring and a linear dashpot this expression becomes

m\ddot{x}+c\dot{x}+kx=0.

(11.2)

We have taken care with the signs of the various terms. Make sure you can confidently derive equation 11.2 without introducing sign errors.

The analytical solution of the damped-oscillator equation is in box 11.1. Some qualitative features of the damped solutions are shown in fig. 11.2

For given $k$ and $m$ we can think of the damping $c$ as adjustable. A system which has small damping (small $c$ ) is under-damped and does not come to equilibrium quickly because oscillations last for a long time. A system which has a lot of damping (big $c$ ) is over-damped and does not come to equilibrium quickly because the dashpot doesn’t leak fast enough. A system which is in between, critically-damped comes to equilibrium most quickly. The purpose of damping is often to purge motion after a disturbance. If the only design variable available for adjustment is the damping, then the quickest purge is accomplished with critical damping, $c=\sqrt{(4km)}$ . In practice, any damping value close to critical is often used, more or less depending on whether a little oscillation is tolerable or not

^†^†margin: Stereotypically, the suspension of an overloaded old-fashioned luxury car is underdamped, imagine it bouncing along after a bump. And the suspension of a tight sports car is overdamped.

Summary of equations for the unforced harmonic oscillator

•

$\ddot{x}+\frac{k}{m}x=0$ , mass-spring equation
•

$\ddot{x}+\lambda^{2}x=0$ , harmonic oscillator equation
•

$x(t)=A\cos(\lambda t)+B\sin(\lambda t)$ , general solution to harmonic oscillator equation
•

$x(t)=R\cos(\lambda t-\phi)$ , amplitude-phase version of solution to harmonic oscillator solution, $R=\sqrt{A^{2}+B^{2}},\phi=\tan^{-1}(\frac{B}{A})$ (See box on page 10.3).
•

$\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x=0$ , mass-spring-dashpot equation (see equations 11.3-11.6 for solutions)

Solution of the damped-oscillator equations

Here are some mathematical details you can use for reference. These details are of much lower status than those in box 3.1 on page 3.1. Only some vibrations experts remember the formulas below in detail.

Even if we make the common assumptions that $m, c,$ and $k$ are all positive, the whole nature of the solution of (11.2) depends on the values of those constants. The three types of solutions are categorized as follows:

•

Under-damped: $c^{2}<4mk$ . In this case the damping is small and oscillations persist forever, though their amplitude diminishes exponentially in time. The general solution for this case is:

$x(t)=e^{(-\frac{c}{2m})t}[A\cos(\lambda_{d}t)+B\sin(\lambda_{d}t)],$ (11.3)

where $\lambda_{d}$ is the damped natural frequency and is given by

$\lambda_{d}=\sqrt{\frac{k}{m}-\left(\frac{c}{2m}\right)^{2}}.$ (11.4)
•

Critically damped: $c^{2}=4mk$ . In this case the damping is at a critical level that separates the cases of under-damped oscillations from the simply decaying motion of the over-damped case. The general solution is:

$x(t)=Ae^{(-\frac{c}{2m})t}+Bte^{(-\frac{c}{2m})t}.$ (11.5)
•

Over-damped: $c^{2}>4mk$ . Here there are no oscillations, just a simple return to equilibrium with at most one crossing through the equilibrium position on the way to equilibrium. The general solution in the over-damped case is:

$\displaystyle x(t)\ =$ $\displaystyle\phantom{+}Ae^{\left(-\frac{c}{2m}+\sqrt{(\frac{c}{2m})^{2}-\frac% {k}{m}}\right)t}$

$\displaystyle+Be^{\left(-\frac{c}{2m}-\sqrt{(\frac{c}{2m})^{2}-\frac{k}{m}}% \right)t}.$ (11.6)

Measurement of damping.

In the under-damped case, the damping constant $c$ can be found by measuring the rate of decay of unforced oscillations using the ‘logarithmic decrement’. The logarithmic decrement is the natural logarithm of the ratio of the amplitude of any two successive peaks. The larger the damping, the greater the rate of decay and the bigger the decrement:

\mbox{logarithmic decrement}\equiv D=\ln(\frac{x_{n}}{x_{n+1}})

(11.7)

where $x_{n}$ and $x_{n+1}$ are the heights of two successive peaks in fig. 11.4 (also seen on the 2nd and 3rd figures in fig. 11.2 on page 11.2).

Because of the exponential envelope (bounding curve), $x_{n}={\rm(const.)}e^{-({c\over 2m})t_{n}}$ and $x_{n+1}={\rm(const.)}e^{-({c\over 2m})(t_{n}+T)}$ .

D=\ln[(e^{-({c\over 2m})t_{n}})/(e^{-({c\over 2m})(t_{n}+T)})]

Simplifying this expression, we get that

D={cT\over 2m}

where $T$ is the period of oscillation. Thus, measuring the logarithmic decrement $D$ and the period of oscillation $T$ determines $c$ as

c={2mD\over T}.

Filename:tfigure8-ang-accel

Figure 11.4: Measuring damping using ‘logarithmic decrements’ (see text).

Solution using complex variables. To find the solutions above it is easiest to use complex variables. Consider this general constant coefficient 2nd order linear equation:

A\ddot{x}+B\dot{x}+Cx=0.

(11.8)

We want to find the most general function $x(t)$ that satisfies this equation. We do this by making the guess that

x(t)=e^{\alpha t}.

Plugging this guess into eqn. (11.8) and cancelling the common non-zero factor $e^{\alpha t}$ from each term gives

A\alpha^{2}+B\alpha+C=0.

For simplicity let’s assume this quadratic has two independent roots $\alpha_{1}$ and $\alpha_{2}$ ,

\alpha_{1,2}=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}.

For each root we have a solution to eqn. (11.8). And the solutions can be multiplied by a constant. And the solutions can be added. Thus the general solution to eqn. (11.8) is:

x(t)=C_{1}e^{\alpha_{1}t}+C_{2}e^{\alpha_{2}t}.

(11.9)

This seems easy enough. But the $\alpha$ ’s might be complex. And the $C$ ’s too. To get a real solution we have to take the real part using the Euler equation $e^{i\lambda t}=\cos\lambda t+i\sin\lambda t$ ). By this means, with lots of algebra, we could see that the solution 11.6 actually includes the solutions 11.3 and 11.5 as special cases. This algebra is carried out for the simplest case, no damping, in box 10.3 on page 10.3.

Numerical solution to the damped oscillator equations

As for the numerical solution of the harmonic oscillator we define $v=\dot{x}$ . Thus

m\ddot{x}+c\dot{x}+kx=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\dot{v}+cv+kx% =0.

Combining the definition of $v$ with the differential equation we get the set of two coupled first order equations

	$\displaystyle\dot{x}$	$\displaystyle=v$
	$\displaystyle\dot{v}$	$\displaystyle=-\frac{k}{m}x-\frac{c}{m}v$		(11.10)

We can think of this as

\dot{z}=f(z)

where $z$ is the list of two numbers $z(1)=x$ and $z(2)=v$ so

\frac{d}{dt}\left[\begin{array}[]{cc}z_{1}\\ z_{2}\end{array}\right]=\left[\begin{array}[]{cc}z_{2}\\ -\frac{k}{m}z_{1}-\frac{c}{m}z_{2}\end{array}\right].

which is standard form for numerical integration.

Energy? Note that for the damped oscillator we cannot use energy conservation to check the solution because energy is constantly lost. We could, however, make a plot of the total energy and make sure that it is an ever decreasing (monotonically decreasing) function of time.

SAMPLE 11.1

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Figure 11.5:

Springs in series versus springs in parallel: Two massless springs with spring constants $k_{1}$ and $k_{2}$ are attached to mass A in parallel (although they look superficially as if they are in series) as shown in Fig. 11.5. An identical pair of springs is attached to mass B in series. Taking $m_{A}=m_{B}=m$ , find and compare the natural frequencies of the two systems. Ignore gravity.

Solution Let us pull each mass downwards by a small vertical distance $y$ and then release. Measuring $y$ to be positive downwards, we can derive the equations of motion for each mass by writing the balance of linear momentum for each as follows.

•

Mass A:

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Figure 11.6: Free-body diagram of the mass.

The free-body diagram of mass A is shown in Fig. 11.6. As the mass is displaced downwards by $y$ , spring 1 gets stretched by $y$ whereas spring 2 gets compressed by $y$ . Therefore, the forces applied by the two springs, $k_{1}y$ and $k_{2}y$ , are in the same direction. The linear momentum balance of mass A in the vertical direction gives:

$\displaystyle\sum F$ $\displaystyle=$ $\displaystyle ma_{y}$

$\displaystyle\hbox to36.135pt{or\hfil}-k_{1}y-k_{2}y$ $\displaystyle=$ $\displaystyle m\ddot{y}$

$\displaystyle\hbox to36.135pt{or\hfil}\ddot{y}+\left(\frac{k_{1}+k_{2}}{m}% \right)y$ $\displaystyle=$ $\displaystyle 0.$

Let the natural frequency of this system be $\omega_{p}$ . Comparing with the standard simple harmonic equation $\ddot{x}+\lambda^{2}x=0$ (see box 3.1 on page 3.1), we get the natural frequency ( $\lambda$ ) of the system:

$\omega_{p}=\sqrt{\frac{k_{1}+k_{2}}{m}}$ (11.11)

Answer: $\omega_{p}=\sqrt{\frac{k_{1}+k_{2}}{m}}$
•

Mass B:

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Figure 11.7: Free-body diagrams

The free-body diagram of mass B and the two springs is shown in Fig. 11.7. In this case both springs stretch as the mass is displaced downwards. Let the net stretch in spring 1 be $y_{1}$ and in spring 2 be $y_{2}$ . $y_{1}$ and $y_{2}$ are unknown, of course, but we know that

$y_{1}+y_{2}=y$ (11.12)

Now, using the free-body diagram of spring 2 and then writing linear momentum balance we get,

$\displaystyle k_{2}y_{2}-k_{1}y_{1}$ $\displaystyle=$ $\displaystyle\underbrace{m}_{0}\;\;a=0$

$\displaystyle y_{1}$ $\displaystyle=$ $\displaystyle\frac{k_{2}}{k_{1}}y_{2}$ (11.13)

Solving (11.12) and (11.13) we get

$y_{2}=\frac{k_{1}}{k_{1}+k_{2}}y.$

Now, linear momentum balance of mass B in the vertical direction gives:

$\displaystyle-k_{2}y_{2}=ma_{y}$ $\displaystyle=$ $\displaystyle m\ddot{y}$

$\displaystyle\hbox to36.135pt{or\hfil}m\ddot{y}+k_{2}\overbrace{\frac{k_{1}}{k% _{1}+k_{2}}y}^{y_{2}}$ $\displaystyle=$ $\displaystyle 0$

$\displaystyle\hbox to36.135pt{or\hfil}\ddot{y}+\frac{k_{1}k_{2}}{m(k_{1}+k_{2}% )}y$ $\displaystyle=$ $\displaystyle 0.$ (11.14)

Let the natural frequency of this system be denoted by $\omega_{s}$ . Then, comparing with the standard simple harmonic equation as in the previous case, we get

$\omega_{s}=\sqrt{\frac{k_{1}k_{2}}{m(k_{1}+k_{2})}}.$ (11.15)

Answer: $\omega_{s}=\sqrt{\frac{k_{1}k_{2}}{m(k_{1}+k_{2})}}$

From (11.11) and (11.15)

$\frac{\omega_{p}}{\omega_{s}}=\frac{k_{1}+k_{2}}{\sqrt{k_{1}k_{2}}}.$

Filename:sfig8-2-3a

Figure 11.8: The ratio $\omega_{p}/\omega_{s}$ of the two natural frequencies as a function of the spring stiffness ratio $k_{1}/k_{2}$ .

Let $k_{1}=k_{2}=k.$ Then, $\omega_{p}/\omega_{s}=2$ , i.e., the natural frequency of the system with two identical springs in parallel is twice as much as that of the system with the same springs in series. Intuitively, the restoring force applied by two springs in parallel will be more than the force applied by identical springs in series. In one case the restoring forces add and in the other they don’t. Therefore, we do expect mass A to oscillate at a faster rate (higher natural frequency) than mass B. The ratio of the two natural frequencies, $\omega_{p}/\omega_{s}$ as a function of $k_{1}/k_{2}$ is plotted in fig. 11.8. As we can see, $\omega_{p}$ is always higher than $\omega_{s}$ for all $k_{1}/k_{2}$ ratios and its minimum value is twice that of $\omega_{s}$ when $k_{1}=k_{2}$ .

Comments:

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Figure 11.9: The two arrangements of springs are equivalent; they are connected in parallel.
1. 1.
  
  Although the springs attached to mass A do not visually seem to be in parallel, from a mechanics point of view they are parallel (see fig. 11.9). You can easily check this result by putting the two springs visually in parallel and then deriving the equation of mass A. You will get the same equations. For springs in parallel, each spring has the same displacement but different forces. For springs in series, each has different displacements (if $k_{1}\neq k_{2}$ ) but the same force.
2. 2.
  
  When many springs are connected to a mass in series or in parallel, sometimes we talk about their effective spring constant, i.e., the spring constant of a single imaginary spring which could be used to replace all the springs attached in parallel or in series. Let the effective spring constant for springs in parallel and in series be represented by $k_{pe}$ and $k_{se}$ respectively. By comparing eqns. (11.11) and (11.15) with the expression for natural frequency of a simple spring mass system, we see that
  
  $k_{pe}=k_{1}+k_{2}\hbox to36.135pt{\hfil and\hfil}\frac{1}{k_{se}}=\frac{1}{k_% {1}}+\frac{1}{k_{2}}.$
  
  These expressions can be easily extended for any arbitrary number of springs, say, $N$ springs:
  
  $k_{pe}=k_{1}+k_{2}+\ldots+k_{N}\hbox to36.135pt{\hfil and\hfil}\frac{1}{k_{se}% }=\frac{1}{k_{1}}+\frac{1}{k_{2}}+\ldots+\frac{1}{k_{N}}.$

SAMPLE 11.2

Figure 11.10 shows two responses obtained from experiments on two spring-mass systems. For each system

1.

Find the natural frequency.
2.

Find the initial conditions.

Filename:sfig8-3-1

Figure 11.10:

Solution

1.

Natural frequency: By definition, the natural frequency $f$ is the number of cycles the system completes in one second. From the given responses we see that:

Case(i):

the system completes $\frac{1}{2}$ a cycle in 1 s.

$\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad f=\frac{1}{2}\mbox{ Hz}.$

Case(ii):

the system completes 1 cycle in 1 s.

$\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad f=1\mbox{ Hz}.$

It is usually hard to measure the fraction of a cycle occurring in a short time. It is easier to first find the time period, i.e., the time taken to complete 1 cycle. ^†^†margin: To estimate the frequency of some repeated motion in an experiment, it is best to measure the time for a large number of cycles, say 5, 10 or 20, and then divide that time by the total number of cycles to get an average value for the time period of oscillation.

Then the natural frequency can be found by the formula $f=\frac{1}{T}$ . From the given responses, we find the time period by estimating the time between two successive peaks (or troughs):

From Figure 11.10 we find that for

Case (i):

$f=\frac{1}{T}=\frac{1}{2\,\mathrm{s}}=\frac{1}{2}\mbox{ Hz},$

Case (ii):

$f=\frac{1}{T}=\frac{1}{1\,\mathrm{s}}=1\mbox{ Hz}$

Answer: $\mbox{case (i) }f=\frac{1}{2}\mbox{ Hz}.\quad\mbox{case (ii) }f=1\mbox{ Hz}.$

Initial conditions: Now we are to find the displacement and velocity at $t=0\,\mathrm{s}$ for each case. Displacement is easy because we are given the displacement plot, so we just read the value at $t=0$ from the plots:

Case (i):

x(0)=0.

Case (ii):

x(0)=1\,\mathrm{cm}.

The velocity (actually the speed) is the time-derivative of the displacement. Therefore, we get the initial velocity from the slope of the displacement curve at $t=0$ .

Case (i):

\dot{x}(0)=\frac{dx}{dt}(t=0)=\frac{\pi\,\mathrm{cm}}{1\,\mathrm{s}}=3.14\,% \mathrm{cm}/\,\mathrm{s}.

Case (ii):

\dot{x}(0)=\frac{dx}{dt}(t=0)=\frac{6\pi\,\mathrm{cm}}{1\,\mathrm{s}}=18.85\,% \mathrm{cm}/\,\mathrm{s}.

Thus the initial conditions are

Answer: $\begin{array}[]{lll}\mbox{Case (i):}&x(0)=0&\dot{x}(0)=3.14\,\mathrm{cm}/\,% \mathrm{s}\\ \mbox{Case (ii):}&x(0)=1\,\mathrm{cm}&\dot{x}(0)=18.85\,\mathrm{cm}/\,\mathrm{% s}\end{array}$

Comments: Estimating the speed from the initial slope of the displacement curve at $t=0$ is not a very good method because it is hard to draw an accurate tangent to the curve at $t=0$ . A slightly different line but still seemingly tangential to the curve at $t=0$ can lead to significant error in the estimated value. A better method, perhaps, is to use the known values of displacement at different points and use the energy method to calculate the initial speed. We show sample calculations for the first system:
Case(i): We know that $x(0)=0$ . Therefore the entire energy at $t=0$ is the kinetic energy $=\frac{1}{2}mv_{0}^{2}$ . At $t=0.5\,\mathrm{s}$ we note that the displacement is maximum, i.e., the speed is zero. Therefore, the entire energy is potential energy $=\frac{1}{2}kx^{2}$ , where $x=x(t=0.5\,\mathrm{s})=1\,\mathrm{cm}$ .

Now, from the conservation of energy:

$\displaystyle\frac{1}{2}mv_{0}^{2}$	$\displaystyle=$	$\displaystyle\frac{1}{2}k(x_{t=0.5\,\mathrm{s}})^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad v_{0}$	$\displaystyle=$	$\displaystyle\sqrt{\frac{k}{m}}\mathord{\cdot}\;(x_{t=0.5\,\mathrm{s}})$
	$\displaystyle=$	$\displaystyle\underbrace{\sqrt{\frac{k}{m}}}_{\lambda}\mathord{\cdot}\;(1\,% \mathrm{cm})$
	$\displaystyle=$	$\displaystyle 2\pi f\mathord{\cdot}(1\,\mathrm{cm})$
	$\displaystyle=$	$\displaystyle 2\pi\mathord{\cdot}\frac{1}{2}\mbox{ Hz}\mathord{\cdot}1\,% \mathrm{cm}$
	$\displaystyle=$	$\displaystyle 3.14\,\mathrm{cm}/\,\mathrm{s}.$

Similar calculations can be done for the second system.

SAMPLE 11.3

Filename:sfig8-3-1a

Figure 11.11: Spring-mass dashpot.

A block of mass $10\,\mathrm{kg}$ is attached to a spring and a dashpot as shown in Figure 11.11. The spring constant $k=1000\,{\mbox{N}}/\,\mathrm{m}$ and a damping rate $c=50\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m}$ . When the block is at a distance $d_{0}$ from the left wall the spring is relaxed. The block is pulled to the right by $0.5\,\mathrm{m}$ and released. Assuming no initial velocity, find

1.

the equation of motion of the block.
2.

the position of the block at $t=2\,\mathrm{s}$ .

Filename:sfig8-5-wiper

Figure 11.12: Free-body diagram of the mass at an instant

t

Solution

1.

Let $x$ be the position of the block, measured positive to the right of the static equilibrium position, at some time $t$ . Let $\dot{x}$ be the corresponding speed. The free-body diagram of the block at the instant $t$ is shown in Figure 11.12.

Since the motion is only horizontal, we can write the linear momentum balance in the $x$ -direction ( $\sum F_{x}=m\,a_{x}$ ):

$\displaystyle\underbrace{-kx-c\dot{x}}_{\sum F_{x}}=m\underbrace{\ddot{x}}_{a_% {x}}$

$\displaystyle\mbox{or}\quad\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x=0$ (11.16)

which is the desired equation of motion of the block.

Answer: $\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x=0.$
2.

To find the position and velocity of the block at any time $t$ we need to solve Eqn (11.16). Since the solution depends on the relative values of $m$ , $k$ , and $c$ , we first compute $c^{2}$ and compare with the critical value $4mk$ .

$\displaystyle c^{2}$ $\displaystyle=$ $\displaystyle 2500(\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m})^{2}$

$\displaystyle\mbox{and}\quad 4mk$ $\displaystyle=$ $\displaystyle 4\mathord{\cdot}10\,\mathrm{kg}\mathord{\cdot}1000\,{\mbox{N}}/% \,\mathrm{m}=4000(\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m})^{2}.$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad c^{2}$ $\displaystyle<$ $\displaystyle 4mk.$

Therefore, the system is underdamped and we may write the general solution as (see box 11.1 on page 11.1)

$x(t)=e^{-\frac{c}{2m}t}\;[A\cos\lambda_{D}t+B\sin\lambda_{D}t]$ (11.17)

where

$\displaystyle\lambda_{D}$ $\displaystyle=$ $\displaystyle\sqrt{\frac{k}{m}-\left(\frac{c}{2m}\right)^{2}}=9.682\,\mathrm{% rad}/\mathrm{s}.$

Substituting the initial conditions $x(0)=0.5\,\mathrm{m}$ and $\dot{x}(0)=0\,\,\mathrm{m}/\mathrm{s}$ in Eqn (11.17) (we need to differentiate Eqn (11.17) first to substitute $\dot{x}(0)$ ), we get

$\displaystyle x(0)$ $\displaystyle=$ $\displaystyle 0.5\,\mathrm{m}=A.$

$\displaystyle\dot{x}(0)$ $\displaystyle=$ $\displaystyle 0=-\frac{c}{2m}\cdot A+\lambda_{D}\mathord{\cdot}B$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad B$ $\displaystyle=$ $\displaystyle\frac{A\,c}{2m\lambda_{D}}=\frac{(0.5\,\mathrm{m})\mathord{\cdot}% (50\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m})}{2\mathord{\cdot}(10% \,\mathrm{kg})\mathord{\cdot}(9.682\,\mathrm{rad}/\mathrm{s})}=0.13\,\mathrm{m}.$

Thus, the solution is

$x(t)=e^{(-2.5\frac{1}{\,\mathrm{s}})t}\left[0.50\cos(9.68\,\mathrm{rad}/% \mathrm{s}\;t)+0.13\sin(9.68\,\mathrm{rad}/\mathrm{s}\;t)\right]\,\mathrm{m}.$

Substituting $t=2\,\mathrm{s}$ in the above expression we get $x(2\,\mathrm{s})=0.003\,\mathrm{m}.$

Answer: $x(2\,\mathrm{s})=0.003\,\mathrm{m}.$

SAMPLE 11.4

A structure, modeled as a single degree of freedom system, exhibits characteristics of an underdamped system under free oscillations. The response of the structure to some initial condition is determined to be $x(t)=Ae^{-\xi\lambda t}\sin(\lambda_{D}t)$ where $A=0.3\,\mathrm{m}$ , $\xi\equiv\mbox{damping ratio}=0.02$ , $\lambda\equiv\mbox{undamped circular frequency}=1\,\mathrm{rad}/\mathrm{s}$ , and $\lambda_{D}\equiv\mbox{damped circular frequency}=\lambda\sqrt{1-\xi^{2}}\approx\lambda$ .

1.

Find an expression for the ratio of energies of the system at the $(n+1)$ th displacement peak and the $n$ th displacement peak.
2.

What percent of energy available at the first peak is lost after 5 cycles?

Solution

1.

We are given that

$x(t)=Ae^{-\xi\lambda t}\sin(\lambda_{D}t).$

The structure attains its first displacement peak when $\sin\lambda_{D}t$ is maximum, i.e.,

$\lambda_{D}t=\frac{\pi}{2}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t=\frac{\pi% }{2\lambda_{D}}.$

At this instant,

$\displaystyle x(t)$ $\displaystyle=$ $\displaystyle Ae^{-\xi\cdot\lambda\cdot\frac{\pi}{2\lambda_{D}}}$

$\displaystyle=$ $\displaystyle Ae^{-\frac{\pi}{2}\cdot\frac{\xi}{\sqrt{1-\xi^{2}}}}$

$\displaystyle=$ $\displaystyle(0.3\,\mathrm{m})\cdot e^{-0.0314}$

$\displaystyle=$ $\displaystyle 0.29\,\mathrm{m}.$

Let $x_{n}$ and $x_{n+1}$ be the values of the displacement at the $n$ th and the $(n+1)$ th peak, respectively. Since $x_{n}$ and $x_{n+1}$ are peak displacements, the respective velocities are zero at these points. Therefore, the energy of the system at these peaks is given by the potential energy stored in the spring. That is

$E_{n}=\frac{1}{2}kx_{n}^{2}\quad\mbox{and}\quad E_{n+1}=\frac{1}{2}kx_{n+1}^{2}.$ (11.18)

Let $t_{n}$ be the time at which the $n$ th peak displacement $x_{n}$ is attained, i.e.,

$x_{n}=Ae^{-\xi\lambda t_{n}}$ (11.19)

Since $x_{n+1}$ is the next peak displacement, it must occur at $t=t_{n}+T_{D}$ where $T_{D}$ is the time period of damped oscillations. Thus

$x_{n+1}=Ae^{-\xi\lambda(t_{n}+T_{D})}$ (11.20)

From Eqns (11.18), (11.19), and (11.20)

$\displaystyle\frac{E_{n+1}}{E_{n}}$ $\displaystyle=$ $\displaystyle\frac{\frac{1}{2}k(Ae^{-\xi\lambda(t_{n}+T_{D}})^{2}}{\frac{1}{2}% k(Ae^{-\xi\lambda t_{n}})^{2}}$

$\displaystyle=$ $\displaystyle e^{-2\xi\lambda T_{D}}.$

Answer: $\frac{E_{n+1}}{E_{n}}=e^{-2\xi\lambda T_{D}}.$
2.

Noting that $T_{D}=\frac{2\pi}{\lambda_{D}}$ and $\lambda_{D}=\lambda\sqrt{1-\xi^{2}}$ , we get

$\displaystyle{E_{n+1}}$ $\displaystyle=$ $\displaystyle{E_{n}}\,e^{-2\xi\cancel{\lambda}\cdot\frac{2\pi}{\cancel{\lambda% }\sqrt{1-\xi^{2}}}}$

$\displaystyle=$ $\displaystyle e^{-4\pi\frac{\xi}{\sqrt{1-\xi^{2}}}}\approx e^{-4\pi\xi}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad E_{n+1}$ $\displaystyle=$ $\displaystyle e^{-4\pi\xi}\,E_{n}.$

Applying this equation recursively for $n=n-1,n-2,\ldots,1,0,$ , we get

$\displaystyle E_{n}$ $\displaystyle=$ $\displaystyle e^{-4\pi\xi}\mathord{\cdot}E_{n-1}$

$\displaystyle=$ $\displaystyle e^{-4\pi\xi}\mathord{\cdot}(e^{-4\pi\xi}\cdot E_{n-2})$

$\displaystyle=$ $\displaystyle(e^{-4\pi\xi})^{3}\mathord{\cdot}E_{n-3}$

$\displaystyle\vdots$

$\displaystyle=$ $\displaystyle(e^{-4\pi\xi})^{n}\mathord{\cdot}E_{0}.$

Now we use this equation to find the percentage of energy of the first peak ( $n=0$ ) lost after 5 cycles ( $n=5$ ):

$\displaystyle\Delta E_{5}$ $\displaystyle=$ $\displaystyle\frac{E_{0}-E_{5}}{E_{0}}\times 100$

$\displaystyle=$ $\displaystyle\left(1-e^{-4\pi\xi\;\mathord{\cdot}\;5}\right)\times 100$

$\displaystyle=$ $\displaystyle 71.5\%.$

Answer: $\Delta E_{5}=71.5\%.$

SAMPLE 11.5 A SDOF spring-mass model from given data: The following table is obtained for successive peaks of displacement from the simulation of free vibration of a mechanical system. Make a single degree of freedom mass-spring-dashpot model of the system choosing appropriate values for mass, spring stiffness, and damping rate.

Data:

peak no. $n$ 0 1 2 3 4 5 6 time ( $\,\mathrm{s}$ ) 0.0000 0.6279 1.2558 1.8837 2.5116 3.1395 3.7674 peak $x$ ( $\,\mathrm{m}$ ) 0.5006 0.4697 0.4411 0.4143 0.3892 0.3659 0.3443

Filename:sfig8-5-wiper-a

Figure 11.13: Oscillation data from the simulation of a mechanical system

Solution Since the data provided is for successive peak displacements, the time between any two successive peaks represents the period of oscillations. It is also clear that the system is underdamped because the successive peaks are decreasing. We can use the logarithmic decrement method to determine the damping in the system.

First, we find the time period $T_{D}$ from which we can determine the damped circular frequency $\lambda_{D}$ . From the given data we find that

t_{2}-t_{1}=t_{3}-t_{2}=t_{4}-t_{3}=\cdots=0.6279\,\mathrm{s}

Therefore,

	$\displaystyle T_{D}$	$\displaystyle=$	$\displaystyle 0.6279\,\mathrm{s}.$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\lambda_{D}$	$\displaystyle=$	$\displaystyle\frac{2\pi}{T_{D}}=10\,\mathrm{rad}/\mathrm{s}.$		(11.21)

Now we make a table for the logarithmic decrement of the peak displacements:

peak disp. $x_{n}\;(\,\mathrm{m})$ 0.5006 0.4697 0.4411 0.4143 0.3892 0.3659 0.3443 $\frac{x_{n}}{x_{n+1}}$ 1.0658 1.0648 1.0647 1.0645 1,0637 1.0627 $\ln\left(\frac{x_{n}}{x_{n+1}}\right)$ 0.0637 0.0628 0.0627 0.0624 0.0618 0.0608

Thus, we get several values of the logarithmic decrement $D=\ln\left(\frac{x_{n}}{x_{n+1}}\right)$ ^†^†margin: Theoretically, all of these values should be the same, but it is rarely the case in practice. When $x_{n}$ ’s are measured from an experimental setup, the values of $D$ may vary even more.

We take the average value of $D$ :

D=\bar{D}=0.0624.

(11.22)

Let the equivalent single degree of freedom model have mass $m$ , spring stiffness $k$ , and damping rate $c$ . Then

\lambda_{D}=\lambda\sqrt{1-\xi^{2}}\approx\lambda=\sqrt{\frac{k}{m}}.

Thus, from Eqn (11.21),

\frac{k}{m}=\lambda^{2}=100(\,\mathrm{rad}/\mathrm{s})^{2},

(11.23)

and, since $D=\frac{cT_{D}}{2m}$ , from Eqn (11.22) we get

$\displaystyle c$	$\displaystyle=$	$\displaystyle\frac{2mD}{T_{D}}$	(11.24)
	$\displaystyle=$	$\displaystyle\frac{2m(0.0624)}{0.6279\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle(0.1988\frac{1}{\,\mathrm{s}})m.$

Equations (11.23) and (11.24) have three unknowns: $k$ , $m$ , and $c$ . We cannot determine all three uniquely from the given information. So, let us pick an arbitrary mass $m=5\,\mathrm{kg}$ . Then

$\displaystyle k$	$\displaystyle=$	$\displaystyle(100\frac{1}{\,\mathrm{s}^{2}})\mathord{\cdot}(5\,\mathrm{kg})$
	$\displaystyle=$	$\displaystyle 500\,{\mbox{N}}/\,\mathrm{m},\quad$
and
$\displaystyle c$	$\displaystyle=$	$\displaystyle(0.1988\frac{1}{\,\mathrm{s}})\mathord{\cdot}(5\,\mathrm{kg})$
	$\displaystyle=$	$\displaystyle 0.99\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m}.$

Answer: $\begin{array}[]{lll}m&=&5\,\mathrm{kg},\\ k&=&500\,{\mbox{N}}/\,\mathrm{m},\\ c&=&0.99\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m}.\end{array}$

Of course, we could choose many other sets of values for $m,\;k$ , and $c$ which would match the given response. In practice, there is usually a little more information available about the system, such as the mass of the system. In that case, we can determine $k$ and $c$ uniquely from the given response.

Problems for 11.1 Free Vibration of a SDOF System

Preparatory Problems

11.1.1 A mass $m$ is connected to a spring $k$ and released from rest with the spring stretched a distance $d$ from its static equilibrium position. It then oscillates back and forth repeatedly crossing the equilibrium. How much time passes from release until the mass moves through the equilibrium position for the second time? Neglect gravity and friction. Answer in terms of some or all of $m$ , $k$ , and $d$ . Answer: Time span $=3\pi\sqrt{m/k}/2$

11.1.2 A spring $k$ with rest length $\ell_{0}$ is attached to a mass $m$ which slides frictionlessly on a horizontal ground as shown. At time $t=0$ the mass is released from rest with the spring stretched a distance $d$ . Measure the mass position $x$ relative to the wall.

(a)

What is the acceleration of the mass just after release?
(b)

Find a differential equation which describes the horizontal motion $x$ of the mass.
(c)

What is the position of the mass at an arbitrary time $t$ ?
(d)

What is the speed of the mass when it passes through $x=\ell_{0}$ (the position where the spring is relaxed)?

Filename:sfig8-5-2disks

Figure 11.14:

11.1.3 Reconsider the spring-mass system from problem 11.14.

(a)

Find the potential and kinetic energy of the spring mass system as functions of time.
(b)

Assigning numerical values to the various variables, use a computer to make a plot of the potential and kinetic energy as a function of time for several periods of oscillation. Are the potential and kinetic energy ever equal at the same time? If so, at what position $x(t)$ ?
(c)

Make a plot of kinetic energy versus potential energy. What is the phase relationship between the kinetic and potential energy?

11.1.4 For the three spring-mass systems shown in the figure, find the equation of motion of the mass in each case. All springs are massless and are shown in their relaxed states. Ignore gravity. (In problem (c) assume vertical motion.) Answer: (a) $m\ddot{x}+kx=F(t)$ , (b) $m\ddot{x}+kx=F(t)$ , and (c) $m\ddot{y}+2ky-2k\ell_{0}\frac{y}{\sqrt{\ell_{0}^{2}+y^{2}}}=F(t)$

Filename:sfig8-4-1

Figure 11.15:

11.1.5 A mass-spring oscillator hangs vertically under gravity. The mass rests in static equilibrium by stretching the spring by an amount $y_{\rm static}=0.025\,\mathrm{m}$ . Take your favorite value of $g$ and find the natural frequency of the oscillator. How much time does the oscillator take to complete one oscillation?

Filename:sfig8-4-1a

Figure 11.16:

11.1.6 A mass moves on a frictionless surface. It is connected to a dashpot with damping coefficient $b$ to its right and a spring with constant $k$ and rest length $\ell$ to its left. At the instant of interest, the mass is moving to the right and the spring is stretched a distance $x$ from its position where the spring is unstretched. There is gravity.

(a)

Draw a free-body diagram of the mass at the instant of interest.
(b)

Derive the equation of motion of the mass. Answer: LHS of Linear Momentum Balance: $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-(kx+b\dot{x})\hat{% \boldsymbol{\imath}}+(N-mg)\hat{\boldsymbol{\jmath}}$ .

Filename:sfig8-6-2

Figure 11.17:

11.1.7 A mass-spring-dashpot system has $m=1\,\mathrm{kg}$ , $k=10\,{\rm kN}/{\rm m}$ , and $c=5\,\mathrm{kg}/\,\mathrm{s}$ . Find the natural frequency, damping ratio, and the damped frequency of the system. Specify whether the system is underdamped, critically damped or overdamped. Answer: $\omega_{n}=100\,\mathrm{rad}/\mathrm{s}$ , $\xi=0.025$ , System is underdamped.

11.1.8 The natural frequency, $\lambda_{n}$ , of a SDOF system is 150 rad/s. Find the minimum damping ( $\xi$ ) that the system must have for the resonant frequency to occur below 100 rad/s? Answer: $\xi\geq 0.74$

More-Involved Problems

11.1.9 The equation of motion of an unforced mass-spring-dashpot system is, $m\ddot{x}+c\dot{x}+kx=0$ , as discussed in the text. For a system with $m=0.4\,\mathrm{kg},c=10\,\mathrm{kg}/\,\mathrm{s}$ , and $k=5\,{\rm N}/{\rm m}$ ,

(a)

Find whether the system is underdamped, critically damped, or overdamped. Answer: Overdamped
(b)

Sketch a typical solution of the system.
(c)

Make an accurate plot of the response of the system (displacement vs time) for the initial conditions $x(0)=0.1\,\mathrm{m}$ and $\dot{x}(0)=0$ .

11.1.10 You are given to design a SDOF damped oscillator that should show no oscillations at all when disturbed from the equilibrium (i.e., it should return to equilibrium without overshooting on the other side). You are given a spring with stiffness $k=500\,{\rm N}/{\rm m}$ , a hydraulic damper with $c=10\,\mathrm{kg}/\,\mathrm{s}$ , and you have a choice of masses from $m=1\,\mathrm{kg}$ to $m=10\,\mathrm{kg}$ in increments of half kg. Find the appropriate mass.

11.1.11 Two SDOF oscillators with the same $k$ and $m$ but different $c$ ’s are hung from the ceiling as shown in the figure. The one on the left is pulled down 2 cm and let go. The other is pulled down by 0.2 cm and let go. Which oscillator undergoes more oscillations before reaching the steady state. Find the steady state displacement of each mass.

Filename:sfig8-6-2a

Figure 11.18:

11.1.12 Experiments conducted on free oscillations of a damped oscillator reveal that the amplitude of oscillations drops to 25% of its peak value in just 3 periods of oscillations. The period of oscillation is measured to be $0.6\,\mathrm{s}$ and the mass of the system is known to be $1.2\,\mathrm{kg}$ . Find the damping coefficient and the spring stiffness of the system. Answer: $k=132\,{\rm N}/{\rm m}$ , $c=1.76\,\mathrm{kg}/\,\mathrm{s}$ .

11.1.13 You are required to design a mass-spring-dashpot system that, if disturbed, returns to its equilibrium position the quickest. You are given a mass, $m=1\,\mathrm{kg}$ , and a damper with $c=10\,\mathrm{kg}/\,\mathrm{s}$ . What should be the stiffness of the spring? Your solution needs to include your definition of “‘quickest”. Answer: $k=25\,{\rm N}/{\rm m}$

11.2 Forcing and resonance

If the world of oscillators was as we have described them so far, especially in Section 10.3, there wouldn’t be much to talk about. The undamped oscillators (of which there are none) would be oscillating away and the damped oscillators (all the real ones) would be damped out to no motion. The reason vibrations exist is because they are somehow excited. This excitement is also called forcing whether or not it is due to a literal mechanical force.

Filename:efig1-2-28

Figure 11.19: A forced mass-spring-dashpot is just a mass held in place by a spring and dashpot but pushed by a force

F(t)

from some external source.

The most important idea of this section is the following

If you shake something at about the same frequency at which it naturally oscillates you will eventually get large motions.

The rest of the section is largely a fleshing out of this idea.

The simplest example of a ‘forced’ harmonic oscillator is the mass-spring-dashpot system with an additional mechanical force applied to the mass. See fig. 11.19. Most of this section will be a study of this system. The governing equation for a forced damped oscillator can be derived from the free-body diagram as follows, where vector notation helps keep the signs right:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle-F_{s}\hat{\boldsymbol{\imath}}-F_{d}\hat{\boldsymbol{\imath}}+F(% t)\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle ma\hat{\boldsymbol{\imath}}$
$\displaystyle\left\{\ \left(-kx-c\dot{x}+F(t)\right)\hat{\boldsymbol{\imath}}\right.$	$\displaystyle=$	$\displaystyle\left.m\ddot{x}\hat{\boldsymbol{\imath}}\ \right\}$
$\displaystyle\left\{\right\}\cdot\hat{\boldsymbol{\imath}}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\quad-kx-c\dot{x}+F(t)$	$\displaystyle=$	$\displaystyle m\ddot{x}$

which is often re-arranged as

m\ddot{x}+c\dot{x}+kx=F(t).

(11.25)

When $F(t)=0$ , there is no forcing and the governing equation reduces to that of the un-forced damped harmonic oscillator, eqn. (11.2).

Equivalent ways to force an oscillator

There are many ways to “force” a system that all lead to the same forced-oscillator equation.

Filename:pfigure-blue-118-2

Figure 11.20: In all cases shown above the same forced oscillator eqn. (11.25) applies. In (a) a literal force is applied. In all the other cases the “forcing” is by a motor that moves something back and forth a distance

\delta

. In (b) the support moves. In (c) and (d) just the spring or just the dashpot end is displaced. In (e) an extra mass is moved relative to the main mass.

1.

With a literal force as in fig. 11.19, shown again in fig. 11.20a.
2.

By shaking the support, as in fig. 11.20b.
3.

By displacing one end of the spring, but not the dashpot as in fig. 11.20c.
4.

By displacing one end of the dashpot, but not the spring as in fig. 11.20d.
5.

By displacing a second mass attached to the first with a motor that controls relative position, as in fig. 11.20e.

That these five systems all lead to the same governing equation follows from drawing free-body diagrams, applying momentum balance, and collecting terms to match the form eqn. (11.25). Note that the meaning of some of the terms in the forced-oscillation equation is different for each system.

Types of forcing

In general this or that machine or structure could be forced in any number of complicated ways. But there are two special forcings of most common engineering interest:

•

$F(t)=F_{0}$ (Constant force), and
•

$F(t)=F\cos\lambda t$ (sinusoidal forcing).

^†^†margin: Note again that we use $\lambda$ (lambda) here instead of the more commonly used $\omega$ (omega) because we want to avoid confusion with the magnitude of angular velocity $\omega=|{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}|$ which itself could have oscillatory motion. It would be confusing (and bad math) to write $\omega=A\sin\omega t$ with $\omega$ having two different meanings in the same equation.

Constant force idealizes situations where the force doesn’t vary much as due say, to gravity, a steady wind, or sliding dry friction. Sinusoidally varying forces are used to approximate oscillating forces as caused, say, by a vibrating support or earthquakes. Forces that are not sinusoidal can be thought of as sums of sine waves thus, in some sense, by knowing how a structure responds to sinusoidal forcing, at various frequencies, you know how it responds to all possible forcings^†^†margin: The best approximation of a function as a sum of sine waves is a Fourier series, a topic you learn in math, advanced physics or linear systems courses. .

Let’s look at each of these two cases in detail.

Forcing with a constant force

The case of constant forcing is both common and easy to analyze, so easy that it is often ignored (see fig. 10.33 on page 10.33).

If $F(t)=F_{0}=\mbox{constant}$ , then the general solution of eqn. (11.25) for $x(t)$ is the same as the unforced case but with a constant added. The constant is $F_{0}/k$ . The usual way of accommodating this case is to describe a new equilibrium point at $x=F_{0}/k$ and to pick a new deflection variable that is zero at that point. If we pick a new variable $z$ defined as $z=x-F_{0}/k$ , then substituting into eqn. (11.25) we get

m\ddot{z}+c\dot{z}+kz=0,

(11.26)

which is the unforced oscillator equation. That is, constant forcing reduces to the case of no forcing if one merely changes what one calls zero to be the place where the mass is in equilibrium, taking account of the spring stretch (or compression) caused by the constant applied force. Thus the solution of the forced equation for $x$ is equivalent to the unforced solution for $z$ :

z(t)=x(t)-F_{0}/k=e^{(-\frac{c}{2m})t}\left(A\cos(\lambda_{\rm d}t)+B\sin(% \lambda_{\rm d}t)\right)

(11.27)

where $\lambda_{\rm d}=\sqrt{\left(\frac{c}{2m}\right)^{2}-k/m}$ , as explained in box 11.1 on page 11.1.

An alternative approach is to use superposition. Here we say $x(t)=x_{h}(t)+x_{p}(t)$ where $x_{h}(t)$ satisfies $m\ddot{x}+c\dot{x}+kx=0$ and $x_{p}(t)$ is any solution $x_{p}$ of $m\ddot{x}+c\dot{x}+kx=F_{0}$ . Any solution you like is called a “particular” solution. One easy solution is $x_{p}=F_{0}/k$ . So the net solution is $x_{p}=F_{0}/k$ plus a solution $x_{h}$ to the ‘homogeneous’ equation 11.26.

x(t)=\underbrace{e^{(-\frac{c}{2m})t}\left(A\cos(\lambda_{\rm d}t)+B\sin(% \lambda_{\rm d}t)\right)}_{x_{h}}+\underbrace{F_{0}/k}_{x_{p1}}

(11.28)

Example: Hanging mass.

Filename:pfigure-s95f3a

Figure 11.21: A mass hangs from a spring and dashpot. Its distance

x

from the support is due to the rest length

\ell_{0}

of the spring, and the stretch of the spring.

The mass hanging from the support shown in fig. 11.21 obeys the equation

m\ddot{x}+c\dot{x}+kx=\underbrace{k\ell_{0}+mg}_{F_{0}}

One particular solution $x_{p}$ , the easiest one, has the mass hanging still. In this solution, the mass position is the un-stretched length $\ell_{0}$ of the spring plus the stretch of the spring due to gravity, $\Delta x=mg/k$ . Because the mass is still in this solution, the dashpot constant $c$ doesn’t appear. So

x_{p}=\ell_{0}+mg/k.

The homogeneous solution $x_{h}$ is given by (11.27) and the general motion is the sum

	$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle x_{p}+x_{h}$
		$\displaystyle=$	$\displaystyle\left(\ell_{0}+mg/k\right)+e^{-\frac{ct}{2m}}\left(C\cos(\lambda_% {\rm d}t)+D\sin(\lambda_{\rm d}t)\right)$

where $C$ and $D$ are constants determined by the initial conditions. For any initial condition and corresponding values of $C$ and $D$ , the motion eventually decays to the stationary particular solution with the mass hanging still (because the exponentials go to zero as $t\rightarrow\infty$ ).

Forcing with a sinusoidally varying force

The motion resulting from sinusoidal forcing is of central interest in vibration analysis. In this case we imagine that $F(t)=F\cos pt$ where $F$ is the amplitude of forcing and $p$ is the angular frequency of the forcing. Note, we could just as well use $F(t)=F\sin pt$ for the forcing, $\sin$ and $\cos$ are both sinusoidal forcings.

The general solution of equation 11.25 is given by the sum of two parts. One is the general solution of equation 11.2, $x_{h}(t)$ , and the other is any solution of equation 11.25, $x_{p}(t)$ . The solution $x_{h}(t)$ of the damped oscillator equation 11.2 is called the ‘homogeneous’ or ‘complementary’ solution. Any solution $x_{p}(t)$ of the forced oscillator equation 11.25 is called a ‘particular’ solution.

We already know the solution $x_{h}(t)$ of the damped governing differential equation 11.2. This solution is equation 11.3, 11.5, or 11.6, depending on the values of the mass, spring and damping constants. So the new problem is to find any solution to the forced equation 11.25. The easiest way to solve this (or any other) differential equation is to make a fortuitous guess (you may learn other methods in your math classes). In this case with

F(t)=F\cos(pt)

we make the guess that

x_{p}(t)=A\cos(pt)+B\sin(pt).

(11.29)

Basically this guess says “If you shake something with a sine wave it will probably move as a sine wave. But who knows the amplitude or phase?” Plugging this guess into the forced oscillator equation (11.25) we find values for $A$ and $B$ in box 11.2 on page 11.2.

Alternatively, a sum of sine waves can be written as a cosine wave (or sine wave) that has been shifted in phase as (see box 10.3 on page 10.3)

x_{p}(t)=A_{0}\cos(pt-\phi),

The value of forced amplitude is simply $A_{0}=\sqrt{A^{2}+B^{2}}$ . This is given in terms of $m, c, k, p$ and $F$ in box 11.2.

The forced amplitude $A_{0}$ is the central subject of this section. It answers the question ‘How big are the oscillations when you shake something?’ Because the formula for $A_{0}$ is admittedly a mess, the answer is often given in a plot

^†^†margin: Another more important reason that a plot is used is that often in a physical system one can measure the vibrations while never knowing a detailed accurate set of differential equations which would describe the system accurately.

The general solution, therefore, is

x(t)=x_{h}(t)+x_{p}(t).

(11.30)

The homogeneous solution $x_{h}(t)$ , the motion of the unforced system, is just decaying oscillations and is usually not of primary interest in vibrating systems. The particular solution $x_{p}(t)$ is steady oscillations. These oscillations are of central interest. In particular most often in engineering one wants these oscillations to be big or small.

Example: MEMs devices.

One general type of “Micro Electronic Machine” consists of, basically, a vibrating beam. A beam with an effective mass $50\mu\,\hbox{gm}$ and effective stiffness of $k=500\,{\rm N}/{\rm m}=5\mu\,{\mbox{N}}/\mu\,\mathrm{m}$ has

\lambda_{\rm n}=\sqrt{k\over m}=\sqrt{\frac{5\,{\rm N}/{\rm m}}{50\cdot 10^{-9% }\,\mathrm{kg}}}=\sqrt{\frac{\frac{500{\rm N}}{\,\mathrm{m}}\left(\frac{1\,% \mathrm{kg}\,\mathrm{m}/s^{2}}{1{\rm N}}\right)}{50\cdot 10^{-6}\,\mathrm{kg}}% }=\sqrt{10^{10}}\,\mathrm{s}^{-1}=10^{5}/\,\mathrm{s}

which corresponds to a frequency of $\lambda_{\rm n}/2\pi\approx 15.7{\rm k}\,\,{\mathrm{hz}}$ . That is, such a MEMs device would be a good receiver (or ‘resonator’) for $15.7{\rm k}\,\,{\mathrm{hz}}$ ultra-sonic vibrations. In this case resonance is useful to make the sensor sensitive.

The size of the oscillations scales with the size of the forcing $F$ (this proportionality is known as ‘linearity’) and also depends on all the parameters $m, c, k$ and $p$ .

Frequency response and resonance

One way to show a structure’s sensitivity to oscillatory loads is by a frequency response curve fig. 11.22. One curve shows the amplitude of vibrations vs the forcing frequency. The main idea of this section, resonance, shows as a peak in the frequency-response curve near the natural frequency $\lambda_{\rm n}=\sqrt{k/m}$ .

Filename:Danef94s1q2

Figure 11.22: Amplitude of oscillation vs forcing frequency for various dampings. Each curve shows the gain

G

vs the forcing frequency for a fixed damping. Note that when the damping is small (

c\ll 1

) and the forcing is close to the natural frequency of vibration

p\approx\lambda_{\rm n}

there is a ‘resonant’ peak in the amplitude of the response. The smaller the damping the higher and narrower is this peak. For very high damping the peak is at a slightly lower frequency. The mass-spring-dashpot system shown was used to generate the plots using the formulas from box 11.2 on page 11.2.

Recall that the natural frequency $\lambda_{\rm n}$ is the unforced frequency of undamped oscillation. The damped natural frequency $\lambda_{\rm d}$ , the frequency of decaying oscillations with damping present, is slightly slower (see, e.g., eqn. (11.3) on page 11.3.). The resonant frequency $\lambda_{\rm res}$ , the frequency of forcing for which the amplitude of motion is maximum (eqn. (11.2) on page 11.2), is slightly lower still.

But, especially when the damping is low, there is only a small difference between the natural frequency, the damped frequency and the resonant frequency. So, in common language and engineering practice they are usually treated as one and the same.

In summary, the frequency response curve has a peak with forcing near to, but not exactly at, the natural frequency of unforced and undamped motion. But most engineers can reasonably assume, even though it’s not exactly true, that resonance occurs when the forcing frequency is the natural vibration frequency.

Resonance is good and bad

Sometimes an engineer studies vibrations with the hope of minimizing them, sometimes with the hope of maximizing them. Resonance is sometimes the problem and sometimes the solution.

Resonant vibrations are usually undesirable in machinery or cars. The vibrations can lead to large stresses, undesirable motions, or unpleasant sounds. A building resonating to earthquake vibrations may be more likely to fall down.

On the other hand, nuclear Magnetic Resonance imaging is used for medical diagnosis. In the old days, the resonant excitation of a clock pendulum was used to keep time. The resonance of quartz crystals is used to time most watches now-a-days. Self excited resonance is what makes musical instruments have such clear pitches. And resonant vibrations are used to give a larger signal in micro-mechanical sensors. In the electrical domain, radio tuners depend on resonance to pick out just one radio band.

Other systems

Most machines and structures are not exactly a point mass moving in one direction and constrained by a single spring and single dashpot. On the other hand, almost all machines have mass, elastic give, and some dissipation when they move. So most machines have natural oscillations after they are banged or disturbed somehow. And so most structures and machines can be shaken to large motions if the appropriate (or inappropriate, depending on your aims) frequency of force is applied

^†^†margin: Golden Gate bridge cables. On a walk a few years ago one of us noted that the vertical cables on the Golden Gate bridge could be induced to oscillate quite visibly if pushed by a person at the right frequency (about 0.5 hz). One cable, maybe 100 tons of steel, was nicely going back and forth about a half a meter. A police car pulled up and stopped. Through the megaphone the officer authoritatively and sternly threatened “If you break it you have to pay for it.”

So the concepts introduced here for a single mass-spring-dashpot system apply to much more complex machines and structures. In particular, they have natural vibration frequencies and they shake a lot (resonate) if forced at near those frequencies.

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Figure 11.23: Natural, damped and resonant frequencies. The upper dashed plot shows the ratio of damped to natural frequency vs forcing frequency ratio

\xi

. The solid curve shows how the resonant frequency varies from the natural frequency as a function of frequency ratio. The lower plot shows how the amplification

G=A_{0}/(F/k)

is nearly the same at the natural, damped and resonant frequencies. That the peak in the

c=0.6{\rm kg/s}

curve of fig. 11.22 is slightly to the left of

p=\lambda_{\rm n}=\sqrt{k/m}

shows as

\lambda_{\rm res}/\lambda_{\rm n}

being slightly below 1 for

\xi\approx 0.3

. For low damping an engineer can treat the natural, damped and resonant frequencies as equal. Similarly the amplification when forced at the natural frequency is very close to the amplification when forcing is at the damped or at the resonant frequency.

Box 11.1 A Loudspeaker cone is a forced oscillator.

Filename:203II-2

Cross-sectional view of a speaker.

A speaker, similar to the ones used in many home and auto speaker systems, is one of many devices which may be conveniently modeled as a one-degree-of-freedom mass-spring-dashpot system. A typical speaker has a paper or plastic cone, supported at the edges by a roll of plastic foam (the surround), and guided at the center by a cloth bellows (the spider). It has a large magnet structure, and (not visible from outside) a coil of wire attached to the point of the cone, which can slide up and down inside the magnet. (The device described above is, strictly speaking, the speaker driver. A complete speaker system includes an enclosure, one or more drivers, and various electronic components.) When you turn on your stereo, the amplifier forces a current through the coil in time with the music, causing the coil to alternately attract and repel the magnet. This rapid oscillation of attraction and repulsion results in the vibration of the cone which you hear as sound.

In the speaker, the primary mass is comprised of the coil and cone, though the air near the cone also contributes as ‘added mass.’ The ‘spring’ and ‘dashpot’ effects in the system are due to the foam and cloth supporting the cone, and perhaps to various magnetic effects. Speaker system design is greatly complicated by the fact that the air surrounding the speaker must also be taken into account. Changing the shape of the speaker enclosure can change the effective values of all three mass-spring-dashpot parameters. (You may be able to observe this dependence by cupping your hands over a speaker (gently, without touching the moving parts), and observing amplitude or tone changes.) Nevertheless, knowledge of the basic characteristics of a speaker (e.g., resonance frequency), is invaluable in speaker system design.

Our approximate equation of motion for the speaker is identical to that of the ideal mass-spring-dashpot above, even though the forcing is from an electromagnetic force, rather than a direct mechanical force:

m\ddot{x}+c\dot{x}+kx=F(t)\hbox{ with }\ F(t)=\alpha i(t)

(11.31)

where $i(t)$ is the electrical current flow through the coil in amps, and $\alpha$ is the electro-mechanical coupling coefficient, in force per unit current.

Experimental measurement

Because no real thing of interest is exactly a single mass-spring-dashpot the ideas of vibrations analysis are often not expressed in terms of ( $m, c$ and $k$ ). Rather, the more broad ideas of natural frequency, frequency response, and resonance are considered on their own.

Using either a large-scale computer model (say a ‘finite-element’ model) or measurement of the physical system itself, one can draw a frequency-response curve like fig. 11.22 on page 11.22.

Here’s how. First you apply a sinusoidal force to the structure at the point of interest, say $F=F\cos(pt)$ . Then you measure the motion of a part of the structure of interest. You might instead measure a strain or rotation, but for definiteness let’s assume you measure the displacement of some point on the structure $\delta$ .

If the structure is linear and has some damping, the eventual motion of the structure will eventually be a sinusoidal oscillation. In particular, you will measure that

\delta=A_{0}\cdot\cos(pt-\phi).

(11.32)

If you had applied half as big a force, you would have measured half the displacement, still assuming the structure is linear, so the ratio of the displacement to the force $A_{0}/F$ is independent of the size of the force $F$ . Let’s define:

G=\frac{A_{0}}{F}\\

That is, the amplification gain $G$ is the ratio of the amplitude of the displacement sine wave to the amplitude of the forcing sine wave. Plotting $p$ on the $x$ axis and $G$ on the $y$ axis, this experiment gives one point on the frequency response curve. Repeating for a range of forcing frequencies one can plot up the frequency response $G=G(p)$ .

Example: Shake table for earthquake response.

One way to get a frequency response curve for a building is to put a scale model on a “shake table”. The base is then moved sinusoidally through a range of frequencies and the motion of the model is observed. This way one can find peaks in the frequency-response curve. These are frequencies that, to the extent they are prevalent in a feared earthquake, are likely to cause damage.

Transient response

As discussed, the full solution of eqn. (11.25) with forcing $F(t)=F_{0}+F\cos pt$ is the sum of three terms

x(t)=x_{h}+x_{p1}+x_{p2}

The first of these has decaying oscillations, the second is a constant, and the third has steady oscillations. When added up the motion can look quite complicated, as seen in fig. 11.24d. The main point is that after some initial complicated transient the motion eventually decays to steady oscillations ( $x_{p2}(t)=A_{0}(\cos pt-\phi)$ ) plus an offset ( $x_{p1}=F/k$ ).

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Figure 11.24: Transient response. (a) shows a suddenly applied force

F_{0}

. The response (b) to this force is a motion that starts at the initial position

x_{0}

(

x_{0}>0

in this illustration) and then oscillates about the new equilibrium

F_{0}/k

. The motion is identical to unforced motion, but offset. Thus it can be used to evaluate the rate of decay of oscillation and the damped period of oscillation. (c) A sinusoidal forcing causes. (d) the response if the mass is released at

x=x_{0}

and suddenly both a constant force

F

and a sinusoidal force

F\sin pt

are applied. The motion eventually settles into a sinusoidal oscillation at the forcing frequency (which is a little longer period than the damped oscillation in this illustration) with amplitude

A_{0}

The vocabulary of forced oscillations

Forced oscillations are so important and common that there is a specialized vocabulary for many of the terms and collections of commonly appearing terms. Here is a list, starting with the terms you know well.

: $m=$ the mass of the particle that is oscillating. For more complicated systems the mass $m$ may represent an “effective” or “equivalent” mass.
: $c=$ the damping coefficient. $c$ is used to describe the viscous drag, the resistance to motion $F_{d}=-c\dot{x}$ .
: $k=$ the spring constant. $k$ describes the elastic restoring “spring” force $F_{s}=-kx$ .
: $F=$ the forcing amplitude for a sinusoidally varying applied force $F(t)=F\sin pt$ or $F(t)=F\cos pt$ or $F(t)=A\sin pt+B\cos pt$ with $F\equiv\sqrt{A^{2}+B^{2}}$ .
: $p=$ the forcing frequency. Some books will use the symbol $\omega$ for the forcing frequency.

The rest of the quantities below are completely determined by the quantities above ( $m, c, k, F$ and $p$ ).

$\lambda_{\rm n}\equiv\sqrt{k/m}$ is the natural frequency. This is the frequency of oscillation if there is neither forcing nor damping. In that case $x(t)=A\cos\lambda_{\rm n}t+B\sin\lambda_{\rm n}t$ . Many books use $\omega_{n}$ for the natural frequency.

$c_{\rm crit}=2*\sqrt{km}$ is the critical damping coefficient. The relation of the actual damping $c$ to the critical damping $c_{\rm crit}$ tells you whether a system is over-damped ( $c>c_{\rm crit}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}$ decay to equilibrium, when unforced, that is exponential) or under-damped ( $c<c_{\rm crit}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}$ decay to equilibrium, when unforced, that is oscillatory). See Fig. 11.2 on page 11.2. Sometimes $c_{\rm crit}$ is more simply written as $c_{c}$ or $c_{cr}$ .

$\xi\equiv c/c_{\rm crit}$ is the damping ratio. The single number $\xi$ (‘ksee’) tells you if a system is over damped ( $\xi>1$ ) or underdamped ( $\xi<1$ ).

$r\equiv p/\lambda_{\rm n}=p/\sqrt{k/m}$ is the frequency ratio . If $r>1$ then the forcing is faster than the frequency of natural unforced vibrations. If $r<1$ then the forcing is slower than the natural vibrations.

$A_{0}=$ the response amplitude . When a steady oscillatory force is applied the motion is eventually oscillatory. The amplitude of the motions is $A_{0}$ , as in $x=A_{0}cos(pt-\phi)$ with

A_{0}=(F/k)/\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}.

$G\equiv A_{0}/(F/k)$ is the gain or amplification. $G$ is the ratio of the eventual amplitude of the oscillator to the response that would occur if the same force was applied at zero frequency. It is the response amplitude scaled by the displacement that would occur if the same force was applied to a spring.

$\lambda_{\rm res}=\lambda_{\rm n}\sqrt{1-2\xi^{2}}$ is the resonant frequency. $\lambda_{\rm res}$ (also called $\lambda_{r}$ or $\omega_{r}$ ) is the frequency such that if $p=\lambda_{\rm res}$ the amplification gain $G$ is maximum. The resonant frequency is the frequency at which you force a system to get the biggest motions. The resonant frequency $\lambda_{\rm res}$ is rather close to the natural frequency $\lambda_{\rm n}$ in systems with small damping ratios. And these are also the systems that are prone to resonant vibrations.

$\lambda_{\rm d}$ is the damped natural frequency . If an underdamped system is released from rest it oscillates as the motions decay. The frequency of these oscillations is

\lambda_{\rm d}=\lambda_{\rm n}\sqrt{1-\xi^{2}}.

The frequency $\lambda_{\rm d}$ of damped oscillations is a shade slower than the frequency $\lambda_{\rm n}$ of oscillation of the same system with no damping. When damping is small the natural frequency $\lambda_{\rm n}$ , the damped frequency $\lambda_{\rm d}$ and the resonant frequency $\lambda_{r}$ are all close to each other (See fig. 11.22a).

$G_{\rm n},G_{\rm res}$ & $G_{\rm d}$ are the amplification gains (see $G$ above) when forcing is at the natural, the resonant and the damped natural frequency respectively ( $p=\lambda_{\rm n},\lambda_{\rm res}$ & $\lambda_{\rm d}$ , see above). $G_{\rm res}$ is the biggest of these by definition. But it is not actually much bigger than $G_{n}$ or $G_{d}$ . These gains can be calculated using the formulas for $G$ and $A_{0}$ above. They are plotted on fig. 11.22b.

$D$ is the logarithmic decrement. $D$ measures the rate of decay of unforced ( $F=0$ ) oscillations. The experimental definition, derivable from a graph of the motion, is

D=\ln(\frac{x_{n}}{x_{n+1}}).

In terms of $m$ , $c$ and $k$ the logarithmic decrement is $D={cT\over 2m}=2\pi\xi$ , as derived on page 11.1. If there is little damping, $c$ is small ( $\xi\ll 1$ ) and $D\approx(x_{n}-x_{n+1})/x_{n}$ is the fractional decrease in amplitude per oscillation. If $D=.1$ then each oscillation is about 10% smaller in amplitude than the previous one.

$Q$ is the quality factor. For the mass-spring-dashpot system it is another way of describing the rate of decay of unforced oscillations.

	$\displaystyle Q$	$\displaystyle\equiv 2\pi(\mbox{energy of oscillator})/(\mbox{energy lost per % cycle})$
		$\displaystyle=2\pi x_{n}^{2}/(x_{n}^{2}-x_{n+1}^{2})$
		$\displaystyle\approx\pi/D=1/(2\xi)\quad\mbox{(for small damping)}\par$

The $\pi$ in the definition of $Q$ makes it so there is no $\pi$ in the formula for the quality factor $Q$ in terms of the damping ratio $\xi$ . Note that, so long as damping is small, $\xi,D$ and $Q$ can each be found approximately from the other. A system with low damping ( $\xi\ll 1$ ) has high quality ( $Q\gg 1$ ) and slowly decaying oscillations and hence a small logarithmic decrement ( $D\ll 1$ ).

Box 11.2 Solution of the forced oscillator equation

The main equation for understanding forced oscillations is:

m\ddot{x}+c\dot{x}+kx=F_{0}+F\cos pt.

Because the equation is linear we look for a solution which is the sum of three terms

x(t)=x_{h}+x_{p1}+x_{p2}

where $x_{h}$ is the homogeneous solution from Eqns 11.3 - 11.6 on page 11.3, depending on whether the system is underdamped (oscillatory decay), critically damped or over damped (non-oscillatory exponential decay). $x_{p1}$ is a particular solution for the constant forcing $F_{0}$ . $x_{p1}(t)$ was found in eqn. (11.28) on page 11.28 to be, simply, $x_{p1}=F_{0}/k$ .

The last part of the solution, finding an $x_{p2}$ for the forcing term $F\cos pt$ is found by guessing

x_{p2}=A\cos pt+B\sin pt.

When this guess is plugged into the equation

m\ddot{x}+c\dot{x}+kx=F\cos pt

every term is either a multiple of $\sin pt$ or $\cos pt$ . Thus we get

\{\mbox{A collection of constants}\}\cos pt+\{\mbox{Another collection}\}\sin pt=0

The only way a sum of a sine wave and cosine wave can be zero for all time is for both coefficients to be zero. Setting the two collections of constants above both to zero gives two simultaneous equations for the unknowns $A$ and $B$ in terms of $m, c, k$ and $p$ . These can be solved to give

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\frac{(F/k)\left(1-\frac{p^{2}}{\left(k/m\right)}\right)}{\left(% \frac{c^{2}}{km}\right)\left(\frac{p^{2}}{k/m}\right)+\left(1-\frac{p^{2}}{% \left(k/m\right)}\right)^{2}},$
	$\displaystyle\mbox{and}\quad\quad B$	$\displaystyle=$	$\displaystyle\frac{(F/k)(cp/k)}{\left(\frac{c^{2}}{km}\right)\left(\frac{p^{2}% }{k/m}\right)+\left(1-\frac{p^{2}}{\left(k/m\right)}\right)^{2}}\quad.$

So we have found the particular solution for forcing with $F(t)=F\cos pt$ , using $A$ and $B$ above, as

\par x_{p2}=A\cos pt+B\sin pt.\par

(11.33)

An alternative form for the solution is

x_{p2}(t)=A_{0}\cos(pt-\phi),

(11.34)

for which we can find the constants $A_{0}$ and $\phi$ using the trig identity $\cos(\theta-\phi)=\cos\theta\cos\phi+\sin\theta\sin\phi$ described in box 10.3 on page 10.3. Applying this identity to the solution above we find the object of central interest, the forced amplitude

$\displaystyle A_{0}$	$\displaystyle=$	$\displaystyle\sqrt{(A^{2}+B^{2})}=\frac{F/k}{\sqrt{\left(\frac{c^{2}}{km}% \right)\left(\frac{p^{2}}{k/m}\right)+\left(1-\frac{p^{2}}{\left(k/m\right)}% \right)^{2}}}$
and also the phase angle
$\displaystyle\phi$	$\displaystyle=$	$\displaystyle\tan^{-1}\left(\frac{B}{A}\right)=\tan^{-1}\left(\frac{cp/k}{% \left(1-\frac{p^{2}}{\left(k/m\right)}\right)}\right).$	(11.36)

All of the expressions above can be somewhat simplified if write them in terms of the frequency ratio $r=p/\lambda_{\rm n}=p/\sqrt{k/m}$ and damping ratio $\xi=c/c_{\rm crit}=c/2\sqrt{km}$ (The frequency ratio, damping ratio and some more specialized vibration words are defined on page 11.2.). Using these dimensionless quantities, the values of the constants in the solution $x_{p2}(t)$ , namely eqn. (11.33) or eqn. (11.34), are:

$\displaystyle A$	$\displaystyle=$	$\displaystyle\frac{(F/k)(1-r^{2})}{4\xi^{2}r^{2}+\left(1-r^{2}\right)^{2}},$
$\displaystyle\quad\quad B$	$\displaystyle=$	$\displaystyle\frac{(F/k)\left(2\xi r\right)}{4\xi^{2}r^{2}+\left(1-r^{2}\right% )^{2}}\quad,$
$\displaystyle A_{0}$	$\displaystyle=$	$\displaystyle\sqrt{(A^{2}+B^{2})}=\frac{{F}/{k}}{\sqrt{(2\xi r)^{2}+\left(1-r^% {2}\right)^{2}}},$
and
$\displaystyle\phi$	$\displaystyle=$	$\displaystyle\tan^{-1}\left(\frac{B}{A}\right)=\tan^{-1}\left(\frac{2\xi r}{1-% r^{2}}\right).$

These constants are for the particular forced solution $x_{p2}(t)$ of eqn. (11.33) or eqn. (11.34).

Again, most important in all of this is the amplitude $A_{0}$ of the forced response. As you can see, the bottom of the fraction for $A_{0}$ gets quite small for small damping ( $\xi\ll 1$ ) if the frequency ratio $r$ is close to 1. That is,

the amplitude is big if the forcing is close to the natural frequency.

Resonant frequency

In detail, the frequency at which the vibration amplitude $A_{0}$ is maximum is not exactly the unforced undamped natural frequency $\lambda_{\rm n}=\sqrt{k/m}$ . The resonant frequency $\lambda_{\rm res}$ is found by maximizing $A_{0}$ with respect to $r=\lambda/\lambda_{\rm n}$ . Setting $dA_{0}/dr=0$ and solving for $r$ we find

r_{\rm res}=\sqrt{1-2\xi^{2}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\lambda_% {\rm res}=\lambda_{\rm n}\sqrt{1-2\xi^{2}}

(11.37)

The ratio of $\lambda_{\rm res}/\lambda_{\rm n}$ is plotted on fig. 11.22 on page 11.22. Also plotted is the ratio of $A_{0}$ at resonance to $A_{0}$ if forcing is at the natural frequency. The morals are that a) for small damping the natural frequency and resonant frequency are very close, and b) for all dampings, there is little error in calculating the amplitude of the maximum vibration response by approximating resonance as being at the natural frequency. Even when resonance is barely a viable concept, for systems that are critically damped, the error is only 40%.

Similarly one might think the damped natural frequency

\par\lambda_{\rm d}=\lambda_{\rm n}\sqrt{1-\xi^{2}}\par\par

would be a better approximation to the resonant frequency. Actually,it’s about half way between the natural and resonant frequencies, as can be seen also on fig. 11.22.

SAMPLE 11.6

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Figure 11.25:

The mass-spring-dashpot system shown in the figure consists of a mass $m=2\,\mathrm{kg}$ , a spring with stiffness $k=3200\,{\rm N}/{\rm m}$ and a dashpot with damping coefficient $c=10\,\mathrm{kg}/\,\mathrm{s}$ .

1.

Is the system underdamped, critically damped or overdamped?
2.

Find the damped natural frequency of the system.
3.

What is the resonant frequency of the system.

Solution

1.

The question about underdamped, critically damped, or overdamped can be answered conveniently by computing the damping ratio $\xi$ . For an underdamped system, $\xi<1$ , for a critically damped system, $\xi=1$ , and for an overdamped system $\xi>1$ . So, let us compute $\xi$ . We know that

$\xi=\frac{c}{c_{c}}=\frac{c}{2\sqrt{km}}.$

Thus, for the given system,

$\xi=\frac{10\,\mathrm{kg}/\,\mathrm{s}}{2\sqrt{3200\,{\rm N}/{\rm m}\cdot 2\,% \mathrm{kg}}}=\frac{10\,\mathrm{kg}/\,\mathrm{s}}{160\,\mathrm{kg}/\,\mathrm{s% }}=0.062.$

Since $\xi<1$ , the system is underdamped.

Answer: Underdamped ( $\xi=0.062$ )
2.

The damped natural frequency, $\lambda_{d}$ , is given by

$\lambda_{d}=\lambda_{n}\sqrt{1-\xi^{2}}$

where $\lambda_{n}=\sqrt{k/m}$ is the natural frequency of the system. Substituting the known values, we get

$\lambda_{d}=\sqrt{\frac{3200\,{\rm N}/{\rm m}}{2\,\mathrm{kg}}}\sqrt{1-(0.062)% ^{2}}=39.92\,\mathrm{rad}/\mathrm{s}$

which is almost the same as the natural frequency $\lambda_{n}=40\,\mathrm{rad}/\mathrm{s}$ .

Answer: $\lambda_{d}=39.92\,\mathrm{rad}/\mathrm{s}$
3.

The resonant frequency of the system, $\lambda_{r}$ , is given by

$\lambda_{r}=\lambda_{n}\sqrt{1-2\xi^{2}}.$

Substituting the known values of $\lambda_{n}$ and $\xi$ , we get

$\lambda_{r}=39.85\,\mathrm{rad}/\mathrm{s}$

which is the smallest among the three characteristic frequencies of the system — natural frequency, damped natural frequency, and the resonant frequency. For small values of $\xi$ , however, the three frequencies are practically indistinguishable as is the case here.

Answer: $\lambda_{r}=39.85\,\mathrm{rad}/\mathrm{s}$

SAMPLE 11.7 Response to a constant force: A constant force $F=50\,{\mbox{N}}$ acts on a mass-spring system as shown in the figure. Let $m=5\,\mathrm{kg}$ and $k=10\,{\rm kN}/\,\mathrm{m}$ .

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Figure 11.26:

1.

Write the equation of motion of the system.
2.

If the system starts from the initial displacement $x_{0}=0.01\,\mathrm{m}$ with zero velocity, find the displacement of the mass as a function of time.
3.

Plot the response (displacement) of the system against time and describe how it is different from the unforced response of the system.

Solution

1.

The free-body diagram of the mass is shown in fig. 11.27 at a displacement $x$ (assumed positive to the right). Applying linear momentum balance in the $x$ -direction, i.e., $(\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}})\cdot\hat{\boldsymbol{\imath}}$ , we get

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Figure 11.27: Free-body diagram of the mass.

$\displaystyle F-kx$ $\displaystyle=$ $\displaystyle m\ddot{x}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\ddot{x}+kx$ $\displaystyle=$ $\displaystyle F$ (11.38)

which is the equation of motion of the system.

The equation of motion has a non-zero right hand side. Thus, it is a nonhomogeneous differential equation. A general solution of this equation is made up of two parts — the homogeneous solution $x_{h}$ which is the solution of the unforced system (eqn. (11.38) with $F=0$ ), and a particular solution $x_{p}$ that satisfies the nonhomogeneous equation. Thus,

x(t)=x_{h}(t)+x_{p}(t).

(11.39)

Now, let us find $x_{h}(t)$ and $x_{p}(t)$ .

Homogeneous solution:

$x_{h}(t)$ has to satisfy the homogeneous equation

m\ddot{x}+kx=0.

Let $\lambda=\sqrt{k/m}$ . Then, from the solution of unforced harmonic oscillator, we know that

x_{h}(t)=A\sin(\lambda t)+B\cos(\lambda t)

where $A$ and $B$ are constants to be determined later from initial conditions.

Particular solution:

$x_{p}$ must satisfy eqn. (11.38). Since the nonhomogeneous part of the equation is a constant ( $F$ ), we guess that $x_{p}$ must be a constant too (of the same form as $F$ ). Let $x_{p}=C$ . Now we substitute $x_{p}=C$ in eqn. (11.38) and solve the resulting equation to determine $C$ :

m\underbrace{\ddot{C}}_{0}+kC=F\hbox to36.135pt{\hfil$\Rightarrow$\hfil}C=F/k% \hbox to36.135pt{\hfil or\hfil}x_{p}=F/k.

Substituting $x_{h}$ and $x_{p}$ in eqn. (11.39), we get

x(t)=A\sin(\lambda t)+B\cos(\lambda t)+F/k.

(11.40)

Now we use the given initial conditions to determine $A$ and $B$ .

$\displaystyle x(t=0)$	$\displaystyle=$	$\displaystyle B+F/k=x_{0}\;(\mbox{given})\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}B=x_{0}-F/k$
$\displaystyle\dot{x}(t)$	$\displaystyle=$	$\displaystyle A\lambda\cos(\lambda t)-B\lambda\sin(\lambda t)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{x}(t=0)$	$\displaystyle=$	$\displaystyle A=0\;(\mbox{given})\hbox to36.135pt{\hfil$\Rightarrow$\hfil}A=0.$
Thus,
$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle(x_{0}-F/k)\cos(\lambda t)+F/k,$	(11.41)
$\displaystyle\hbox to36.135pt{and\hfil}\dot{x}(t)$	$\displaystyle=$	$\displaystyle-\lambda(x_{0}-F/k)\sin(\lambda t).$	(11.42)

3.

Let us plug the given numerical values, $k=10\,{\rm kN}/\,\mathrm{m},m=5\,\mathrm{kg}$ , (which gives $\lambda=\sqrt{k/m}=44.72\,\mathrm{rad}/\mathrm{s}$ ), $F=50\,{\mbox{N}}\mbox{ and }x_{0}=0.01\,\mathrm{m}$ in eqn. (11.41) and (11.42). The displacement and the velocity are now given as

$\displaystyle x(t)$ $\displaystyle=$ $\displaystyle(0.005\,\mathrm{m})\cos(44.72\,\mathrm{rad}/\mathrm{s}\cdot t)+0.% 005\,\mathrm{m},$

$\displaystyle\hbox to36.135pt{and\hfil}\dot{x}(t)$ $\displaystyle=$ $\displaystyle-(0.22\,\mathrm{m}/\mathrm{s})\sin(44.72\,\mathrm{rad}/\mathrm{s}% \cdot t).$

This response is plotted in fig. 11.28 against time. Note that the oscillations of the mass are about a non-zero mean value, $x_{\rm eq}=0.005\,\mathrm{m}$ . A little thought should reveal that this is what we should expect. When a mass hangs from a spring under gravity, the spring elongates a little, by $mg/k$ to be precise, to balance the mass. Thus, the new static equilibrium position is not at the relaxed length $\ell_{0}$ of the spring but at $\ell_{0}+mg/k$ . Any oscillations of the mass will be about this new equilibrium. The velocity, however, has a zero mean value which is what we expect from eqn. (11.42).

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Figure 11.28: Displacement of the mass as a function of time. Note that the mass oscillates about a non-zero value of $x$ .

This problem is exactly like a mass hanging from a spring under gravity, a constant force, but just rotated by $90^{\circ}$ . The new static equilibrium is at $x_{\rm eq}=F/k$ and any oscillations of the mass have to be around this new equilibrium.

We can rewrite the response of the system by measuring the displacement of the mass from the new equilibrium. Let $\tilde{x}=x-F/k$ . Then, eqn. (11.41) becomes

$\tilde{x}=\tilde{x}_{0}\cos(\lambda t)$

where $\tilde{x}_{0}=x_{0}-F/k$ is the initial displacement. Clearly, this is the response of an unforced harmonic oscillator. Thus the effect of a constant force on a spring-mass system is just a shift in its static equilibrium position.

Answer:

SAMPLE 11.8

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Figure 11.29:

A single degree of freedom damped oscillator has unknown mass, spring stiffness and damping coefficient. In order to find these quantities, the oscillator is subjected to a constant force $F_{0}=100\,{\mbox{N}}$ and its transient response is recorded. The response is shown in fig. 11.29. The two peaks marked in the response plot correspond to ( $t, x$ ) = (0.2107 s, 0.01345 m) and (0.3525 s, 0.0117 m) respectively. Find the system parameters $m$ , $k$ , and $c$ .

Solution

Let the mass, stiffness, and damping coefficient of the system be $m$ , $k$ , and $c$ , respectively. Then the equation of motion of the system, subjected to a constant force $F_{0}$ is,

m\ddot{x}+c\dot{x}+kx=F_{0}

where $x(t)$ is the displacement at some instant t. From the solution of this equation, we know that the steady state solution (after the transient oscillations die) is merely a shift in the static equilibrium position, given by $F_{0}/k$ . From the given response, we see that

\frac{F_{0}}{k}=0.01\,\mathrm{m}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}k=% \frac{F_{0}}{0.01\,\mathrm{m}}=\frac{100\,{\mbox{N}}}{0.01\,\mathrm{m}}=10\,{% \rm kN}/\,\mathrm{m}.

Thus we have found one of the parameters, $k$ . Now we need to find $m$ and $c$ .

Since two successive peaks are given in the transient response, we can use the logarithmic decrement to determine the damping ratio $\xi$ from the relationship

\xi=\frac{1}{2\pi}\ln\left(\frac{x_{n}}{x_{n+1}}\right).

From the given data, $x_{n}=0.01345\,\mathrm{m}$ and $x_{n+1}=0.0117\,\mathrm{m}$ . Therefore,

^†^†margin: The calculation here is illustrative. In practice, you should not use just two data points for computing damping from the logarithmic decrement method. In general, we use several data points and determine the damping from taking an average of the decrements. See Sample 11.12 on page 11.12.

\xi=\frac{1}{2\pi}\ln\left(\frac{0.01345\,\mathrm{m}}{0.0117\,\mathrm{m}}% \right)=0.022.

Since $\xi=c/c_{c}=c/(2\sqrt{km})$ , we have

c=2\xi\sqrt{km}=0.044\sqrt{km}

(11.43)

This is just one equation in two unknowns, $m$ and $c$ (we already know $k$ ). So, we need another equation. From the peak to peak distance (in time), we can find the damped time period. That is $T_{d}=T_{2}-T_{1}=0.3525\,\mathrm{s}-0.2107\,\mathrm{s}=0.1418\,\mathrm{s}$ . But, $T_{d}=2\pi/\lambda_{d}$ , and $\lambda_{d}=\lambda_{n}\sqrt{1-\xi^{2}}$ . Therefore,

$\displaystyle\lambda_{n}^{2}\equiv\frac{k}{m}$	$\displaystyle=$	$\displaystyle\frac{\lambda_{d}^{2}}{1-\xi^{2}}=\frac{4\pi^{2}}{T_{d}^{2}(1-\xi% ^{2})}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m$	$\displaystyle=$	$\displaystyle\frac{kT_{d}^{2}(1-\xi^{2})}{4\pi^{2}}$
	$\displaystyle=$	$\displaystyle\frac{10000\,{\rm N}/{\rm m}\cdot(0.1418\,\mathrm{s})^{2}(1-0.022% ^{2})}{4\pi^{2}}$
	$\displaystyle=$	$\displaystyle 5.09\,\mathrm{kg}.$

Now substituting the value of $m$ and $k$ in eqn. (11.43), we get

c=0.044\sqrt{10000\,{\rm N}/{\rm m}\cdot 5.09\,\mathrm{kg}}=9.92\,\mathrm{kg}/% \,\mathrm{s}.

Answer: $m=5.09\,\mathrm{kg},\quad k=10\,{\rm kN}/\,\mathrm{m},\mbox{ and }c=9.92\,% \mathrm{kg}/\,\mathrm{s}.$

SAMPLE 11.9 Damping and forced response: When a single-degree-of-freedom damped oscillator (mass-spring-dashpot system) is subjected to a periodic forcing $F(t)=F_{0}\sin(pt)$ , then the response of the system is given by

x(t)=X\cos(pt-\phi)

where $X=\frac{F_{0}/k}{\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}},\;\phi=\tan^{-1}\frac{2\xi r% }{1-r^{2}},\;r=\frac{p}{\lambda},\;\lambda=\sqrt{k/m}$ and $\xi$ is the damping ratio.

1.

For $r\ll 1$ , i.e., the forcing frequency $p$ much smaller than the natural frequency $\lambda$ , how does the damping ratio $\xi$ affect the response amplitude $X$ and the phase $\phi$ ?
2.

For $r\gg 1$ , i.e., the forcing frequency $p$ much larger than the natural frequency $\lambda$ , how does the damping ratio $\xi$ affect the response amplitude $X$ and the phase $\phi$ ?

Solution

1.

If the frequency ratio $r\ll 1$ , then $r^{2}$ will be even smaller; so we can ignore $r^{2}$ terms with respect to 1 in the expressions for $X$ and $\phi$ . Thus, for $r\ll 1$ ,

Filename:bikefork-alt

Figure 11.30:

$\displaystyle X$ $\displaystyle=$ $\displaystyle\frac{F_{0}/k}{\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}}\approx\frac{F_{% 0}/k}{1}=\frac{F_{0}}{k}$

$\displaystyle\phi$ $\displaystyle=$ $\displaystyle\tan^{-1}(2\xi r)\approx\tan^{-1}0=0$

that is, the response amplitude does not vary with the damping ratio $\xi$ , and the phase also remains constant at zero. As an example, we use the full expressions for $X$ and $\phi$ for plotting them against $\xi$ for $r=0.01$ in fig. 11.30

Answer: $\mbox{For }r\ll 1,X\approx F_{0}/k,\;\mbox{and }\;\phi\approx 0$
2.

If $r\gg 1$ , then the denominator in the expression for $X$ , $4\xi^{2}r^{2}+(1-r^{2})^{2}\approx r^{4}$ (because we can ignore all other terms with respect to $r^{4}$ . Similarly, we can ignore 1 with respect to $r^{2}$ in the expression for $\phi$ . Thus, for $r\gg 1$ ,

Filename:pfigure-blue-128-1

Figure 11.31:

$\displaystyle X$ $\displaystyle=$ $\displaystyle\frac{F_{0}/k}{\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}}\approx\frac{F_{% 0}/k}{r^{2}}=0$

$\displaystyle\phi$ $\displaystyle=$ $\displaystyle\tan^{-1}\frac{2\xi r}{-r^{2}}\approx\tan^{-1}\frac{2\xi}{-r}% \approx\tan^{-1}(-0)=\pi.$

Once again, we see that the response amplitude and phase do not vary with $\xi$ . This is also evident from fig. 11.31 where we plot $X$ and $\phi$ using their full expressions for $r=10$ . The slight variation in $\phi$ around $\pi$ goes away as we take higher values of $r$ .

Answer: $\mbox{For }r\gg 1,X\approx 0,\;\mbox{and }\;\phi\approx\pi$

Thus, we see that the damping in a system does not affect the response of the system much if the forcing frequency is far away from the natural frequency.

SAMPLE 11.10

Refer to caption — Figure 11.32: A MEMS cantilever resonator: Such resonators are typically silicon cantilever beams fabricated using micromachining process. These beams have geometric dimensions in micrometers and can be as small as a couple of micrometers long and a few nanometers in width and thickness.

A MEMS (microelectromechanical system) cantilever resonator (shown in the figure) is modeled as a single degree of freedom oscillator (SDOF) oscillator. Using load deflection measurements, the stiffness of the beam (equivalent to the spring stiffness) is found to be $90\,{\rm N}/{\rm m}$ . The beam is excited using electrical actuation and its resonant frequency is determined under two different conditions: (i) the beam vibrating in vacuum where the viscous damping is negligible, and (ii) the beam vibrating in ambient conditions where the airflow around it causes viscous damping. If the two frequencies are found to be 30 kHz and 28.4 kHz, respectively, find the equivalent mass $m$ and the damping ratio for the SDOF model. If the beam is subjected to a periodic actuation at the free end by a force $F(t)=F\sin(2\pi ft)$ where $F=50\mu\,{\mbox{N}}$ and $f=25$ kHz, find the steady state displacement amplitude and the phase of the free end of the resonator.

Solution

First we need to find $m$ and $c$ for the equivalent mass-spring-dashpot model. In the first case, where the resonant frequency is found in vacuum, we neglect damping, i.e., $c=0$ . Therefore, the given frequency is the natural frequency. However, it is $f_{n}$ , not the circular natural frequency $\lambda_{n}$ . Now, $\lambda_{n}=2\pi f_{n}$ , hence

\sqrt{\frac{k}{m}}=2\pi f_{n}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m=\frac{% k}{4\pi^{2}f_{n}^{2}}=\frac{90\,{\rm N}/{\rm m}}{4\pi^{2}(30000\,\mathrm{s}^{-% 1})^{2}}=2.533\times 10^{-9}\,\mathrm{kg}.

We now use the damped natural frequency to find the damping ratio $\xi$ . Since, we are given $f_{d}=28.4$ kHz, and we know that $\lambda_{d}=\lambda_{n}\sqrt{(}1-\xi^{2})$ , we have

$\displaystyle 2\pi f_{d}$	$\displaystyle=$	$\displaystyle 2\pi f_{n}\sqrt{1-\xi^{2}}$
$\displaystyle\xi$	$\displaystyle=$	$\displaystyle\sqrt{1-\left(\frac{f_{d}}{f_{n}}\right)^{2}}=\sqrt{1-\left(\frac% {28.4\mbox{ kHz}}{30\mbox{ kHz}}\right)^{2}}$
	$\displaystyle=$	$\displaystyle 0.32.$

Now, we know the values of all system parameters for our SDOF model of the MEMS resonator — $m$ , $k$ and $\xi$ (can find $c$ if required from $\xi$ , $k$ and $m$ ). For the given sinusoidal forcing, the equation of motion of the SDOF oscillator is:

m\ddot{x}+c\dot{x}+kx=F\sin(2\pi ft).

We can write the steady state solution as the particular solution $x(t)=A_{0}\sin(pt-\phi)$ where $p=2\pi f$ , and the displacement amplitude $A_{0}$ and the phase $\phi$ are given by the following expressions:

A_{0}=\frac{F/k}{\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}},\hbox to36.135pt{\hfil and% \hfil}\phi=\tan^{-1}\left(\frac{2\xi r}{1-r^{2}}\right).

Since, $r=\frac{p}{\lambda_{n}}=\frac{2\pi f}{2\pi f_{n}}=\frac{25}{30}=0.833$ , we have,

$\displaystyle A_{0}$	$\displaystyle=$	$\displaystyle\frac{(50^{-6}\,{\mbox{N}})/(90\,{\rm N}/{\rm m})}{\sqrt{(2\cdot 0% .32\cdot 0.833)^{2}+(1-0.833^{2})^{2}}}$
	$\displaystyle=$	$\displaystyle 9.04\times 10^{-7}\,\mathrm{m}=0.904\mu{\rm m}.$
Similarly, we find the phase as,
$\displaystyle\phi$	$\displaystyle=$	$\displaystyle\tan^{-1}\frac{2\cdot 0.32\cdot 0.833}{1-0.833^{2}}$
	$\displaystyle=$	$\displaystyle 1.05\,{\rm rad}.$

Answer: $m=2.533\times 10^{-9}\,\mathrm{kg},\xi=0.32,A_{0}=0.904\mu{\rm m},\mbox{ and }% \phi=1.05\,{\rm rad}$

SAMPLE 11.11 Energetics of resonance: Consider the response of a damped harmonic oscillator to a periodic forcing. Find the work done on the system by the periodic force during a single cycle of the force and show how this work varies with the forcing frequency and the damping ratio.

Solution Let us consider the damped harmonic oscillator shown in fig. 11.33 with $F(t)=F\sin(pt)$ . The equation of motion of the system is $m\ddot{x}+c\dot{x}+kx=F\sin(pt)$ and the response of the system may be expressed as $X\sin(pt-\phi)$ where $X=(F/k)/\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}$ and $\phi=\tan^{-1}({2\xi r}/(1-r^{2})),$ with $r=p/\lambda_{n},\;\lambda_{n}=\sqrt{k/m}$ and $\xi=c/(2\sqrt{km})$ .

Filename:sfig8-7-2

Figure 11.33:

We can compute the work done by the applied force on the system in one cycle by evaluating the integral

$\displaystyle W$	$\displaystyle=$	$\displaystyle\int_{\rm onecycle}F(t)\,dx$
But, $x=X\sin(pt-\phi)\hbox to36.135pt{\hfil$\Rightarrow$\hfil}dx=Xp\cos(pt-\phi)dt$ . Therefore,
$\displaystyle W$	$\displaystyle=$	$\displaystyle\int_{0}^{2\pi/p}F\sin(pt)\cdot Xp\cos(pt-\phi)\,dt$
	$\displaystyle=$	$\displaystyle FXp\int_{0}^{2\pi/p}\sin(pt)\cos(pt-\phi)\,dt$
	$\displaystyle=$	$\displaystyle FXp\int_{0}^{2\pi/p}\sin(pt)\left(\cos(pt)\cos\phi+\sin(pt)\sin% \phi\right)\,dt$
	$\displaystyle=$	$\displaystyle FXp\left[\cos\phi\cdot\frac{1}{2}\int_{0}^{2\pi/p}\sin(2pt)\,dt+% \sin\phi\cdot\frac{1}{2}\int_{0}^{2\pi/p}(1-\cos(2pt))\,dt\right]$
	$\displaystyle=$	$\displaystyle\frac{FXp}{2}\left[\cos\phi\left.\left(-\frac{\cos(2pt)}{2p}% \right)\right\|_{0}^{2\pi/p}+\sin\phi\left.\left(t-\frac{\sin(2pt)}{2p}\right)% \right\|_{0}^{2\pi/p}\right]$
	$\displaystyle=$	$\displaystyle\frac{FXp}{2}\left[\frac{\cos\phi}{2p}(-1+1)+\frac{2\pi}{p}\sin% \phi+0\right]$
	$\displaystyle=$	$\displaystyle\frac{FXp}{2}\cdot\frac{2\pi}{p}\sin\phi$
	$\displaystyle=$	$\displaystyle F\pi X\sin\phi$

Although the expression obtained above for $W$ looks simple, we must substitute for $X$ and $\phi$ to see the dependence of $W$ on the damping ratio $\xi$ and the frequency ratio $r$ .

W=\frac{(F\pi)\cdot(F/k)}{\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}}\cdot\frac{2\xi r}% {\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}}=\frac{(2\pi F^{2}\xi r)/k}{(2\xi r)^{2}+(1% -r^{2})^{2}}

(11.44)

Filename:sfig8-7-2a

Figure 11.34:

Unfortunately, this expression is too complicated to see the dependence of $W$ on $\xi$ and $r$ . However, we know that for small $r(\ll 1)$ , $\phi\approx 0$ and for large $r(\gg 1)$ , $\phi\approx\pi$ , implying that $W$ is almost zero in both these cases. On the other hand, for $r$ close to one, that is, close to resonance, $\phi\approx\pi/2\Rightarrow\sin\phi\approx 1$ , but the response amplitude $X$ is large (for small $\xi$ ), which makes $W$ to be big near the resonance. Figure 11.34 shows a plot of $W$ against $r$ , using eqn. (11.44), for different values of $\xi$ . It is clear from the plot that the work done on the system in a single cycle is much larger close to the resonance for lightly damped systems. This explains why the response amplitude keeps on growing near resonance.

Answer: $W=\pi FX\sin\phi$

Problems for 11.2 Forced Vibration and resonance

Preparatory Problems

11.2.1 Given that $\ddot{\theta}+k^{2}\theta=\beta\sin\omega t$ , $\theta(0)=0$ , and $\dot{\theta}(0)=\dot{\theta}_{0}$ , find $\theta(t)$ .

Answer: $\theta(t)=\frac{1}{k}\left(\dot{\theta}_{0}-\frac{\beta\omega}{k^{2}-\omega^{2% }}\right)\sin kt+\frac{\beta}{k^{2}-\omega^{2}}\sin kt$

11.2.2 Three SDOF systems, each with the same mass but different stiffnesses, $k_{1}=k$ , $k_{2}=2k$ , and $k_{3}=4k$ , and different damping, $c_{1}=c$ , $c_{2}=2c$ , and $c_{3}=c/2$ , are subjected to the same periodic forcing, $F=F_{0}\sin pt$ where $p$ is less than the resonant frequency of each of the systems. Sketch approximately the response of each of the three oscillators assuming all of them to be underdamped. Clearly mark the transient and steady state part of the response, and indicate the relative values of the response amplitudes.

11.2.3 A $3\,\mathrm{kg}$ mass is suspended by a spring ( $k=10\,{\rm N}/{\rm m}$ ) and forced by a $5\,{\mbox{N}}$ sinusoidally oscillating force with a period of $1\,\mathrm{s}$ . What is the amplitude of the steady-state oscillations (ignore the “homogeneous” solution) Answer: $A=0.17\,\mathrm{m}$

More-Involved Problems

11.2.4 A machine that can be modeled as a SDOF system is put under vibration test for estimating the system parameters $m$ , $k$ , and $c$ . First, a transient test is conducted by disturbing the machine from its equilibrium and letting it settle down to equilibrium again. The transient response is recorded as a displacement versus time plot and is shown in fig. 11.35(a). Next, a sinusoidal forcing of amplitude $F_{0}$ and angular frequency $p$ is applied on the machine and its steady state response is recorded along with the forcing function. This response is shown in fig. 11.35(b).

(a)

Mark the relevant points on the transient response plot and explain, with equations, which system parameters can be determined using what information from this plot.
(b)

On the steady state plot, mark the phase difference between the response and the forcing function. From the given phase, can you find out whether $p>\lambda_{n}$ or $p<\lambda_{n}$ ?
(c)

From the phase difference of the steady state response and the information obtained from the transient response, can you determine the frequency ratio $r$ ? Explain with appropriate equations.
(d)

From the amplitude of the steady state response, and the rest of the information obtained above, find the rest of the system parameters.

Filename:sfig8-7-2again

Figure 11.35:

11.2.5 A machine produces a steady-state vibration due to a forcing function described by $Q(t)=Q_{0}\sin\omega t$ , where $Q_{0}=5000N$ . The machine rests on a circular concrete foundation. The foundation rests on an isotropic, elastic half-space. The equivalent spring constant of the half-space is $k=2,000,000\,{\rm N}/{\rm m}$ and has a damping ratio $d=c/c_{c}=0.125$ . The machine operates at a frequency of $\omega=4$ Hz.

1.

What is the natural frequency of the system?
2.

If the system were undamped, what would the steady-state displacement be?
3.

What is the steady-state displacement given that $d=0.125$ ?
4.

How much additional thickness of concrete should be added to the footing to reduce the damped steady-state amplitude by 50%? (The diameter must be held constant.)

11.2.6 The transient response of an oscillatory system shows exponential decay of the peak displacements at each cycle. The second peak is found to be twice as big as the fifth peak. Find the damping ratio $x i$ for the system. How many cycles does it take for the peak displacement to drop below 5% of the first peak displacement?

11.2.7 A 50 $\,\mathrm{kg}$ engine is mounted on springs with an equivalent single spring stiffness of 1200 $\,{\rm N}/{\rm m}$ . Using various means, enough damping needs to be provided so that any unwanted vibration dies quickly. Assume that this objective is met by dissipating 80% of the available energy in a single cycle of vibration. Find the damping coefficient of the system.

11.2.8 Consider the system shown in the figure. You are given that $m=10\,\mathrm{kg}$ , $k=50\,{\rm N}/{\rm m}$ , and $c=5\,\mathrm{kg}/\,\mathrm{s}$ . A periodic force $F=F_{0}\cos pt$ acts on the system as shown where $F_{0}=25\,{\mbox{N}}$ and $p=2.5\,\mathrm{rad}/\mathrm{s}$ .

(a)

Find the resonant frequency of the system.
(b)

Find the steady state response of the system, specifying the amplitude and phase of the motion.
(c)

What is the displacement amplification ( $G=A/(F_{0}/k))$ ?
(d)

Find the work done by the force on the system in one cycle.
(e)

Find the energy lost to the damper in one cycle.
(f)

Find the quality factor, Q, of the system using the energy calculations.

Filename:sfig8-7-2disks

Figure 11.36:

11.2.9 A MEMS cantilever beam resonator is used for mass measurement of biological molecules by comparing the shift in the resonant frequency of the beam after the test molecule is attached to the free end of the beam. In a SDOF model of the resonator, it is equivalent to finding the difference in the resonant frequency of the system with mass $m$ and $m+\Delta m$ . If the ’effective mass’ of the beam (mass to be used in the SDOF model) is $2.05\times 10^{-15}\,\mathrm{kg}$ , the stiffness is $0.625\,{\rm N}/{\rm m}$ , and the Q of the resonator is 900, find the shift in the resonant peak in Hz when a biological molecule of mass $1.36\times 10^{-21}$ kg is attached to the end of the beam (equivalently to the mass $m$ ).

11.2.10 A damped mass-spring system is subjected to a constant load $F_{0}=50\,{\mbox{N}}$ by ramping the load to the constant level in (a) $t_{1}=2\,\mathrm{s}$ and (b) $t_{2}=10\,\mathrm{s}$ . If the mass of the system $m=1\,\mathrm{kg}$ , the natural frequency $\lambda_{n}=62\,\mathrm{rad}/\mathrm{s}$ , and the damping ratio $\xi=0.2$ , find the difference in the settling time of the system to the steady state between the two given cases.

Filename:sfig8-4-4

Figure 11.37:

11.2.11 An accelerometer is a sensor that is used to measure acceleration of a body. It can be modeled as a single degree of freedom spring-mass-dashpot system that is attached to a body frame as shown in the figure. Assume that the body undergoes vertical motion denoted by $y(t)$ and, as a result, the mass of the accelerometer undergoes vertical motion $z(t)$ relative to the frame. From a measurement of $z(t)$ we want to know if we can determine the acceleration $\ddot{y}$ of the frame. Neglect gravity.

(a)

What is the absolute or inertial acceleration of the mass in terms of $z(t)$ and $y(t)$ and their derivatives?
(b)

Write the equation of motion of the mass in terms of $z$ and $y$ .
(c)

Assume that $y(t)=y_{0}\sin pt$ . What is the magnitude of acceleration of the frame (that is, the peak acceleration of this sine wave)?
(d)

Find the steady state response $z(t)$ of the accelerometer when $y(t)=y_{0}\sin pt$ (a big mess). Plot the amplitude of the response vs the amplitude of the frame acceleration as the frequency $p$ is varied.
(e)

One would like a signal ( $z$ ) that is proportional to acceleration $\ddot{y}$ independent of the frequency of shaking. Show that this requires that the frequency of frame motion $p$ must be much smaller than the natural frequency $\lambda_{n}$ .
(f)

What is the amplitude of the response (the magnitude of $z$ ) of the accelerometer when $p\gg\lambda_{n}$ ?

Filename:sfig8-4-4a

Figure 11.38:

A single degree of freedom spring-mass model of an accelerometer.

11.2.12 Consider the accelerometer described in Problem 11.37. Assume that the frame undergoes a sinusoidal motion given by $y(t)=y_{0}\sin pt$ .

(a)

Find the response $z(t)$ of the accelerometer.
(b)

Given that $m=0.5\,\mathrm{kg}$ , $k=5$ kN/m, and $c=10$ $\,\mathrm{kg}$ /s, find the maximum acceleration that the accelerometer can sense, assuming the accelerometer to work in the frequency range much below its natural frequency (i.e., $p/\lambda_{n}\ll 1$ ). Express your answer in terms of the gravitational acceleration $g$ (it is customary to talk about acceleration of various things in terms of ‘so many $g$ ’s’).

11.3 Normal modes

Filename:sfig8-4-4b

Figure 11.39: A two mass system. We define

x_{1}

and

x_{2}

so that the system is in equilibrium when

x_{1}=x_{2}=0

To read this section you need to know the linear algebra concepts of eigenvalues and eigenvectors.

Systems with many moving parts often move in complicated ways. Consider the two mass system shown in fig. 11.39. By drawing free-body diagrams and writing linear momentum balance for the two masses we can write the equations of motion in matrix form (see eqn. (10.49)) as

[M]\ddot{\boldsymbol{x}}+[K]\boldsymbol{x}=\boldsymbol{0}

where

[M]=\left[\begin{array}[]{cc}m&0\\ 0&m\\ \end{array}\right]\qquad\mbox{and}\qquad[K]=\left[\begin{array}[]{cc}2k&-k\\ -k&2k\\ \end{array}\right].

Example: Complicated motion.

Filename:sfig8-4-4c

Figure 11.40: Motions of the masses from fig. 11.39 for three different initial conditions, all released from rest (

v_{1}=v_{2}=0)

x_{1}=1,\ x_{2}=\phantom{-}0

, b)

x_{1}=1,\ x_{2}=\phantom{-}1

, and c)

x_{1}=1,\ x_{2}=-1

If we put the initial condition

\boldsymbol{x}_{0}=\left[\begin{array}[]{c}1\\ 0\\ \end{array}\right]\par\qquad\mbox{and}\qquad\par\boldsymbol{v}_{0}=\left[% \begin{array}[]{c}0\\ 0\\ \end{array}\right],

we get the motion shown in fig. 11.40a. Both masses move in a complicated way and not synchronously with each other.

On the other hand, all such systems, if started in just the right way, will move in a simple way.

Example: Simple motion: a normal mode.

If we put the initial condition

\boldsymbol{x}_{0}=\left[\begin{array}[]{c}1\\ 1\\ \end{array}\right]\par\qquad\mbox{and}\qquad\par\boldsymbol{v}_{0}=\left[% \begin{array}[]{c}0\\ 0\\ \end{array}\right].

we get the motion shown in fig. 11.40b. Both masses move in a simple sine wave, synchronously and in phase with each other.

That this system has this simple motion is intuitively apparent. If both of the equal masses are displaced equal amounts both have the same restoring force. So both move equal amounts in the ensuing motion. And nothing disturbs this symmetry as time progresses. In fact the frequency of vibration is exactly that of a single spring and mass (with the same $k$ and $m$ ).

A given system can have more than one such simple motion.

Example: Another normal mode.

If we put the initial condition

\boldsymbol{x}_{0}=\left[\begin{array}[]{c}\phantom{-}1\\ -1\\ \end{array}\right]\par\qquad\mbox{and}\qquad\par\boldsymbol{v}_{0}=\left[% \begin{array}[]{c}0\\ 0\\ \end{array}\right].

we get the motion shown in fig. 11.40c. Both masses move in a simple sine wave, synchronously and exactly out of phase with each other. Being exactly out of phase is actually a form of being exactly in phase, but with a negative amplitude.

This motion is also intuitive. Each mass has restoring force of $3k\Delta x$ . One $k$ from a spring at the end and $2k$ because each mass experiences a spring with half the length (and thus twice the stiffness) in the middle (because the middle of the middle spring doesn’t move in this symmetric motion).

The system above is about the simplest for demonstration of normal mode vibrations. But more complicated elastic systems always have such simple normal mode vibrations.

All elastic systems with mass have normal mode vibrations in which all masses

•

have simple harmonic motion
•

with the same frequency as all the other masses, and
•

exactly in (or out) of phase with all of the other masses

Thus the first and second normal modes from fig. 11.40b,c can be written as

\underbrace{\left[\begin{array}[]{c}x_{1}(t)\\ x_{2}(t)\\ \end{array}\right]=\left[\begin{array}[]{c}\cos\lambda_{1}t\\ \cos\lambda_{1}t\\ \end{array}\right]}_{\mbox{First normal mode}}\par\qquad\mbox{and}\qquad\par% \underbrace{\left[\begin{array}[]{c}x_{1}(t)\\ x_{2}(t)\\ \end{array}\right]=\left[\begin{array}[]{c}\phantom{-}\cos\lambda_{2}t\\ -\cos\lambda_{2}t\\ \end{array}\right]}_{\mbox{Second normal mode}}\par

where, by the physical reasoning in the examples we know that $\lambda_{1}=\sqrt{k/m}$ and $\lambda_{2}=\sqrt{3k/m}$ . We could equally well have used the sine function instead of cosine.

Superposition of normal modes

Note that the governing equation (eqn. (11.3)) is ‘linear’ in that the sum of any two solutions is a solution. If we add the two solutions from fig. 11.40b,c we have a solution. And if we divide that sum by two we get a solution. And not just any solution, but the solution in fig. 11.40a. The top curve is the sum of the bottom two divided by two (The curves for $x_{1}(t)$ and $x_{2}(t)$ need to be added separately).

For more complicated systems it is not so easy to guess the normal modes. Most any initial condition will result in a complicated motion. Nonetheless the concept of normal modes applies to any system governed by the system of equations (eqn. (11.3)):

[M]\ddot{\boldsymbol{x}}+[K]\boldsymbol{x}=\boldsymbol{0}.

Any collection of springs and masses connected any which way has normal mode vibrations. And because elastic solids are the continuum equivalent of a collection of springs and masses, the concept applies to all elastic structures. Here are the basic facts

•

An elastic system with $n$ degrees of freedom has $n$ independent normal modes.
•

In each normal mode $i$ all the points move with the same angular frequency $\lambda_{i}$ and exactly in phase.
•

Any motion of the system is a superposition of normal modes (a sum of motions each of which is a normal mode).

Example: Musical instruments

The pitch of a bell is determined by that normal mode of the bell that has the lowest natural frequency. Similarly for violin and piano strings, marimba keys, kettle drums and the air-column in a tuba.

A recipe for finding the normal modes of more complex systems is given in box 11.3 on page 11.3.

Normal modes and single-degree-of-freedom systems

Any complex elastic system has simple normal mode motions. And all motions of the system can be represented as a superposition of normal modes. Hence sometimes we can think of every system as if it is a single degree of freedom system. For example, if a complex elastic system is forced, it will resonate if the frequency of forcing matches any of its normal mode (or natural) frequencies.

The math of, and how to find, normal modes

Consider a collection of $n$ masses connected by springs whose motions are governed by eqn. (11.3)

[M]\ddot{\boldsymbol{x}}+[K]\boldsymbol{x}=\boldsymbol{0},

where the positions of the masses are $\boldsymbol{x}=\boldsymbol{x}(t)=[x_{1}(t),x_{2}(t),...,x_{n}(t)]^{\prime}$ . The matrices $[M]$ and $[K]$ have to do with the masses and the network of springs, respectively. At this point in the book the examples are masses in a line, but the concepts are more general.

How to find a solution. The approach used by the professionals is to guess that there are “normal mode” solutions and then see if they are. A normal mode solution, with all masses moving sinusoidally and synchronously, is

\boldsymbol{x}=\left[\begin{array}[]{c}V_{1}\cos\lambda t\\ V_{2}\cos\lambda t\\ \vdots\end{array}\right]\par=\boldsymbol{V}\cos\lambda t.

Upper case bold $\boldsymbol{V}$ (to distinguish it from lower case velocity) is a list of constants $[V_{1},V_{2},...]^{\prime}$ . We could have used $\sin$ just as well as $\cos$ for our guess.

Now we plug our guess into the governing equations to see if it is a good guess:

	$\displaystyle[M]\ddot{\boldsymbol{x}}+[K]\boldsymbol{x}$	$\displaystyle=\boldsymbol{0}$
	$\displaystyle[M]{d^{2}\over dt^{2}}\left\{\boldsymbol{V}\cos\lambda t\right\}+% [K]\left\{\boldsymbol{V}\cos\lambda t\right\}$	$\displaystyle=\boldsymbol{0}$
	$\displaystyle-\lambda^{2}[M]\boldsymbol{V}\cos\lambda t+[K]\boldsymbol{V}\cos\lambda t$	$\displaystyle=\boldsymbol{0}$
	$\displaystyle\left\{-\lambda^{2}[M]\boldsymbol{V}+[K]\boldsymbol{V}\right\}% \cos\lambda t$	$\displaystyle=\boldsymbol{0}.$

This equation has to hold true for all $t$ therefore the constant column vector inside the brackets $\displaystyle\left\{\right\}$ must be zero:

	$\displaystyle-\lambda^{2}[M]\boldsymbol{V}+[K]\boldsymbol{V}=\boldsymbol{0}$
	$\displaystyle\left[-\lambda^{2}[M]+[K]\right]\boldsymbol{V}=\boldsymbol{0}$

The matrix $[M]$ is usually invertible. If $[M]$ is diagonal its inverse is $[M]$ with each element replaced by its reciprocal. Assuming $[M]^{-1}$ exists we can multiply through by $[M]^{-1}$ to get:

[M]^{-1}[K]\boldsymbol{V}=\lambda^{2}\boldsymbol{V},

where we used that $[M]^{-1}[M]=[1]$ = the identity matrix, and that $[1]\boldsymbol{V}=\boldsymbol{V}$ . Defining the product $[B]=[M]^{-1}[K]$ and substituting we get the classic eigenvalue problem:

[B]\boldsymbol{V}=\lambda^{2}\boldsymbol{V}.

(11.45)

There is a lot to know about eqn. (11.45). It’s a famous equation. Equation (11.45) says that $\boldsymbol{V}$ is a vector that, when multiplied by $[B]$ gives itself back again, multiplied by a constant. For the special vector $\boldsymbol{V}$ , being multiplied by the matrix $[B]$ is equivalent to being multiplied by the scalar $\lambda^{2}$ .

Given $[B]$ there are generally $n$ linear independent eigen vectors $\boldsymbol{V}^{1},\boldsymbol{V}^{2},...\boldsymbol{V}^{n}$ with associated eigen values $\lambda_{1}^{2},\lambda_{2}^{2},...,\lambda_{n}^{2}$ . Note, $[B]$ is generally not symmetric.

In the case of our vibration problem the eigen vectors are called modes or eigen modes or mode shapes or normal modes.

Recipe for finding normal modes

Given the matrices $[M]$ and $[K]$ proceed as follows.

•

Calculate $[B]=[M]^{-1}K$
•
Use a math computer program to find the eigenvalues and eigenvectors of $[B]$ , call these $\boldsymbol{V}^{i}$ and $\lambda_{i}^{2}$ . Usually this is a single command, like:
```
                   eig(B)   
```
•

For each $i$ between 1 and $n$ write each normal mode as $\boldsymbol{x}(t)=\boldsymbol{V}\cos{(\lambda_{i}t)}$ or as $\boldsymbol{x}(t)=\boldsymbol{V}\sin{(\lambda_{i}t)}$

For example, if

[M]=\left[\begin{array}[]{cc}m&0\\ 0&m\\ \end{array}\right]\qquad\mbox{and}\qquad[K]=\left[\begin{array}[]{cc}-2k&k\\ k&-2k\\ \end{array}\right].

then, for any values of $k$ and $m$ , the computer will return for the eigen values and eigenvectors of $[B]=[M]^{-1}[K]$ :

\boldsymbol{V}^{1}=\left[\begin{array}[]{c}1\\ 1\end{array}\right]\mbox{with\ }\lambda_{1}^{2}=k/m\mbox{\ and \ }\par% \boldsymbol{V}^{2}=\left[\begin{array}[]{c}\phantom{-}1\\ -1\end{array}\right]\mbox{with\ }\lambda_{2}^{2}=3k/m

Why are they called ‘normal’ modes?: Another recipe

The math here is relatively advanced, so trust or skip it if you don’t have the needed linear algebra background. In math speak ‘normal’ sometimes means orthogonal. Here is the sense that ‘normal’ modes are orthogonal to each other.

First, the matrix $[M]$ is generally both symmetric, non-singular and even positive definite, so $[M]$

•

Has a symmetric inverse $[M]^{-1}$ with

$[M]^{-1}[M]=[M][M]^{-1}=[{\bf 1}]=[\mbox{identity matrix}],$
•

Has a unique positive definite square root $\sqrt{[M]}$ with

$\sqrt{[M]}\sqrt{[M]}=[M],$
•

Has a unique inverse square root $[M]^{-1/2}$ with

$[M]^{-1/2}[M][M]^{-1/2}=[{\bf 1}].$

First we use $[M]^{-1/2}$ to change coordinates from $\boldsymbol{x}$ to $\boldsymbol{y}$ as

\boldsymbol{x}=[M]^{-1/2}\boldsymbol{y}\quad\mbox{or}\quad\boldsymbol{y}=[M]^{% 1/2}\boldsymbol{x}.

Now we substitute this into the basic vibration equation ( $[M]\ddot{\boldsymbol{x}}+[K]\boldsymbol{x}=\boldsymbol{0}$ ) and pre-multiply the whole equation by $[M]^{-1/2}$ to get

	$\displaystyle\underbrace{[M]^{-1/2}[M][M]^{-1/2}}_{[{1}]}\ddot{\boldsymbol{y}}% +\underbrace{[M]^{-1/2}[K][M]^{-1/2}}_{[{A}]}\boldsymbol{y}$	$\displaystyle=\boldsymbol{0}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\boldsymbol{y}}+[{% A}]\boldsymbol{y}$	$\displaystyle=\boldsymbol{0}.$

Now we look for solutions for $\boldsymbol{y}$ exactly as we did for $\boldsymbol{x}$ before. But, as opposed to $[B]$ , $[A]$ is symmetric. So $[A]$ has $n$ linear independent and mutually orthogonal eigen vectors $\boldsymbol{W}^{1},\boldsymbol{W}^{2},...\boldsymbol{W}^{n}$ with associated eigen values $\lambda_{1}^{2},\lambda_{2}^{2},...,\lambda_{n}^{2}$ . These $\boldsymbol{W}_{i}$ give the $\boldsymbol{V}_{i}$ by $\boldsymbol{V}_{i}=[M]^{-1/2}\boldsymbol{W}_{i}$ . The $\lambda_{i}$ are the same.

These coordinate-changed $\boldsymbol{W}_{i}=[M]^{1/2}\boldsymbol{V}_{i}$ are ‘normal’ (mutually orthogonal) but the more physical $\boldsymbol{V}_{i}$ are not, even though the $\boldsymbol{V}_{i}$ are called ‘normal’ modes. Or you can say that the $\boldsymbol{V}_{i}$ are mutually orthogonal with respect to the weighting $[M]$ , e.g., $\boldsymbol{V}_{2}^{{}^{\prime}}[M]\boldsymbol{V}_{5}=0.$

SAMPLE 11.12

Filename:pfigure-s94h13p2

Figure 11.41: A MEMS vibratory gyroscope: (a) a micrograph of the two-mass structure. Each inertial mass (the plates with holes) is approximately

1\,{\rm mm}\times 1\,{\rm mm}\times 15\mu\,\mathrm{m}

. The beams that hang the two masses act as springs. The comb drives on the left and right side of the two masses are used to drive the two masses to oscillate in the

x

-direction at their resonant frequencies. (b) a two degree of freedom spring-mass model of the driven gyroscope structure (without any damping).

A two mass vibratory MEMS gyroscope: A vibratory MEMS (microelectromechanical system) gyroscope employs two big plates as inertial masses, suspended by thin beams or ‘springs’ as shown in the figure. The two masses are made to vibrate (by electrical actuation) out of phase in the $x$ -direction. Any rotation about the $y$ -direction causes the masses to vibrate out-of-plane due to ‘Coriolis acceleration’ (you will learn about that in later chapters). We will restrict our attention to the planar motion of the gyroscope. A two degree of freedom spring-mass model is shown in the figure where $m=34.5\times 10^{-9}\,\mathrm{kg}$ , $k_{1}=25\,{\rm N}/{\rm m}$ , and $k_{2}=3\,{\rm N}/{\rm m}$ .

1.

Write the equations of motion for the two masses.
2.

For the out of phase motion of the two masses, assume that $x_{1}(t)=-x_{2}(t)=x_{0}\sin\lambda_{n}t$ . Determine the natural frequency $\lambda_{n}$ corresponding to this mode of vibration.

Solution

1.

The free-body diagram of each mass is shown in fig. 11.42. Assuming both $x_{1}$ and $x_{2}$ to be positive in the $x$ -direction, and $x_{2}>x_{1}$ at the instant shown in the figure, we can write the equations of motion using the balance of linear momentum as

$\displaystyle\mbox{Mass A:}\qquad m\ddot{x}_{1}$ $\displaystyle=$ $\displaystyle k_{2}(x_{2}-x_{1})-2k_{1}x_{1}$

$\displaystyle=$ $\displaystyle-(2k_{1}+k_{2})x_{1}+k_{2}x_{2}$

$\displaystyle\mbox{Mass B:}\qquad m\ddot{x}_{2}$ $\displaystyle=$ $\displaystyle-k_{2}(x_{2}-x_{1})-2k_{1}x_{2}$

$\displaystyle=$ $\displaystyle k_{2}x_{1}-(2k_{1}+k_{2})x_{2}.$

These two equations can be also written in a convenient matrix form as

$\left(\begin{array}[]{c}\ddot{x}_{1}\\ \ddot{x}_{2}\end{array}\right)=\frac{1}{m}\left[\begin{array}[]{cc}-(2k_{1}+k_% {2})&k_{2}\\ k_{2}&-(2k_{1}+k_{2})\end{array}\right]\left(\begin{array}[]{c}{x}_{1}\\ {x}_{2}\end{array}\right).$ (11.46)

Filename:pfigure-f93f5

Figure 11.42: Partial free-body diagram of the two masses. Only relevant forces (in the $x$ -direction) are shown in the diagram.

Answer: $m\ddot{x}_{1}=-(2k_{1}+k_{2})x_{1}+k_{2}x_{2},\quad m\ddot{x}_{2}=k_{2}x_{1}-(% 2k_{1}+k_{2})x_{2}$
2.

The out of phase normal mode of vibration of the two masses is such that $x_{1}(t)=x_{0}\sin\lambda_{n}t$ and $x_{2}=-x_{0}\sin\lambda_{n}t$ , i.e., the two masses have out of phase displacements ( $x_{1}=-x_{2}$ ). If we substitute these values of the displacements, we see that both equations turn out to be the same and they give,

$-\lambda_{n}^{2}x_{0}\sin\lambda_{n}t=\frac{1}{m}(-2k_{1}-k_{2}-k_{2})x_{0}% \sin\lambda_{n}t$

from which it follows that,

$\lambda_{n}=\sqrt{\frac{2(k_{1}+k_{2})}{m}}.$

Substituting the given values of $m$ , $k_{1}$ , and $k_{2}$ , we get,

$\lambda_{n}=\sqrt{\frac{2(25+3)\,{\rm N}/{\rm m}}{34.5\times 10^{-9}\,\mathrm{% kg}}}=40.29\times 10^{3}\,\mathrm{rad}/\mathrm{s}.$

Thus the natural frequency corresponding to the out of phase vibration mode is $40.29\times 10^{3}\,\mathrm{rad}/\mathrm{s}$ which corresponds to $f_{n}=\lambda_{n}/2\pi=6.4$ kHz.

Answer: $f_{n}=6.4{\rm kHz}$

SAMPLE 11.13

Filename:pfigure-s94h13p3

Figure 11.43: The two degree of freedom model of a two-mass MEMS vibratory gyroscope. Here,

m=34.5\times 10^{9}\,\mathrm{kg}

34.5\mu{\rm g}

k_{1}=25\,{\rm N}/{\rm m}

, and

k_{2}=3\,{\rm N}/{\rm m}

Normal modes from eigen analysis: Consider the two-mass MEMS gyroscope of Sample 11.3 again. Using the equations of motion derived in Sample 11.3,

1.

Find the natural frequencies and the corresponding normal modes of vibration of the system.
2.

Using initial conditions based on the normal modes, solve the equations of motion numerically and plot $x_{1}(t)$ and $x_{2}(t)$ together for each normal mode. From the plots, show that the time period of oscillation conforms to the natural frequencies found above.

Solution

1.

The equations of motion for the two degree of freedom model were obtained in eqn. (11.46) and are reproduced here:

$\left(\begin{array}[]{c}\ddot{x}_{1}\\ \ddot{x}_{2}\end{array}\right)=\frac{1}{m}\left[\begin{array}[]{cc}-(2k_{1}+k_% {2})&k_{2}\\ k_{2}&-(2k_{1}+k_{2})\end{array}\right]\left(\begin{array}[]{c}{x}_{1}\\ {x}_{2}\end{array}\right).$ (11.47)

Let us assume a normal mode of vibration in the form

$\left(\begin{array}[]{c}{x}_{1}(t)\\ {x}_{2}(t)\end{array}\right)=\left(\begin{array}[]{c}v_{1}\\ v_{2}\end{array}\right)\sin\lambda_{n}t$

where $\lambda_{n}$ is the natural frequency of the system. Substituting this assumed motion in eqn. (11.47) and getting rid of $\sin\lambda_{n}t$ from both sides, we get,

$-\lambda_{n}^{2}\left(\begin{array}[]{c}{v}_{1}\\ {v}_{2}\end{array}\right)=\frac{1}{m}\left[\begin{array}[]{cc}-(2k_{1}+k_{2})&% k_{2}\\ k_{2}&-(2k_{1}+k_{2})\end{array}\right]\left(\begin{array}[]{c}{v}_{1}\\ {v}_{2}\end{array}\right).$

Rearranging this equation a little bit, we can write it as [A] V = $\lambda$ V, the standard eigenvalue problem, where $\lambda(=\lambda_{n}^{2})$ is the eigenvalue of the matrix [A] and V is the corresponding eigenvector. Here,

${\rm\bf A}=\left[\begin{array}[]{cc}(2k_{1}+k_{2})/m&-k_{2}/m\\ -k_{2}/m&(2k_{1}+k_{2})/m\end{array}\right].$

Now, we can go to a computer and find the eigenvalues and eigenvectors of [A]:
```
    m = 34.5*10^(-9),  k1 = 25,  k2 = 3
    A = [(2*k1+k2)/m     -k2/m;
         -k2/m           (2*k1+k2)/m]
    lambda = eigenvalues(A)
    v = eigenvectors(A)
```
By carrying out this computation, using appropriate commands in a computational package, we find the following two eigenvalues and the corresponding two eigenvectors:

$\displaystyle\lambda^{(1)}=1.449\times 10^{9},$ $\displaystyle{\rm\bf V}^{(1)}=\left(\begin{array}[]{c}1\\ 1\end{array}\right);$

$\displaystyle\hbox to36.135pt{\hfil and\hfil}\lambda^{(2)}=1.623\times 10^{9},$ $\displaystyle{\rm\bf V}^{(2)}=\left(\begin{array}[]{r}1\\ -1\end{array}\right).$

Now, since we know that $\lambda=\lambda_{n}^{2}$ , we can find the natural frequencies of our system by taking the square root of the eigenvalues just found. Thus,

$\displaystyle\lambda_{n}^{(1)}=3.807\times 10^{4}\,\mathrm{rad}/\mathrm{s}$ $\Rightarrow$ $\displaystyle f_{n}^{(1)}=\lambda_{n}^{(1)}/2\pi=6.06\;{\rm kHz},$

$\displaystyle\hbox to36.135pt{and\hfil}\lambda_{n}^{(2)}=4.029\times 10^{4}\,% \mathrm{rad}/\mathrm{s}$ $\Rightarrow$ $\displaystyle f_{n}^{(2)}=\lambda_{n}^{(2)}/2\pi=6.41\;{\rm kHz}.$

The corresponding normal modes or mode shapes are given by V⁽¹⁾ and V⁽²⁾ . Please note that the components of an eigenvector are determined relative to each other, that is, the absolute numerical values are not unique, and any multiple of an eigenvector is also an eigenvector. For example, you could find V ${}^{(1)}=[\sqrt{2}\quad\sqrt{2}]^{T}$ , or V ${}^{(1)}=[1/\sqrt{2}\quad 1/\sqrt{2}]^{T}$ .

Answer: $\begin{array}[]{ll}\lambda_{n}^{(1)}=3.807\times 10^{4}\,\mathrm{rad}/\mathrm{% s},&{\rm\bf V}^{(1)}=[1\quad 1]^{T}\\ \lambda_{n}^{(2)}=4.029\times 10^{4}\,\mathrm{rad}/\mathrm{s},&{\rm\bf V}^{(2)% }=[1\quad-1]^{T}\end{array}$
2.

The normal modes thus found indicate that as long as we set the initial conditions for the two masses in the same proportion as one of the mode shapes (eigenvectors), the two masses will vibrate synchronously with the same frequency (corresponding to the chosen mode shape). So, we now simulate the motion of the two masses by solving the equations of motion numerically, using appropriate initial conditions.

We first write the equations of motion as a set of first order equations:

$\displaystyle\dot{x}_{1}$ $\displaystyle=$ $\displaystyle u_{1}$

$\displaystyle\dot{u}_{1}$ $\displaystyle=$ $\displaystyle-\frac{(2k_{1}+k_{2})}{m}x_{1}+\frac{k_{2}}{m}x_{2}$

$\displaystyle\dot{x}_{2}$ $\displaystyle=$ $\displaystyle u_{2}$

$\displaystyle\dot{u}_{2}$ $\displaystyle=$ $\displaystyle\frac{k_{2}}{m}x_{2}-\frac{(2k_{1}+k_{2})}{m}x_{2}.$

For the first mode, we set the initial conditions $x_{1}(0)=x_{2}(0)=1\mu{\rm m}$ corresponding to the first eigenvector V ${}^{(1)}=[1\quad 1]^{T}$ . Now, we are ready to solve the equations numerically.
```
    ODEs = {x1dot = u1,
            u1dot = -(2*k1+k2)/m * x1 + k2/m * x2,
            x2dot = u2,
            u2dot = k2/m * x1 - (2*k1+k2)/m * x2}
    ICs   = {x1(0)=1E-6, u1(0)=0, x2(0)=1E-6, u2(0)=0}
    Set m = 34.5E-9, k1 = 25, k2 = 3,
    Solve ODEs with ICs for t=0 to t=0.6E-3
    Plot x1(t) and x2(t)
```
Note that we are solving the equations for only 0.6 milliseconds, that is, less than a millisecond. This is because we already know that the frequency is very high, roughly about 6 kHz, which means we can get six oscillations in one millisecond. The plot of $x_{1}(t)$ and $x_{2}(t)$ are shown in fig. 11.44. From this plot, we find that the time period of one oscillation is approximately $1.64\times 10^{-4}$ seconds, which gives a frequency of $\lambda_{n}=2\pi/T=3.8\times 10^{4}\,\mathrm{rad}/\mathrm{s}$ .

Filename:p-s96-p3-3

Figure 11.44: Motions of mass A and mass B, $x_{1}(t)$ and $x_{2}(t)$ in the first normal mode. Note that both masses move together (in phase) and oscillate with the same frequency, $\lambda_{n}^{(1)}$ . In this mode, the middle spring, $k_{2}$ plays no role as it remains unstretched at all times $x_{2}(t)-x_{1}(t)=0$ for all $t$ .

Similarly, using the initial conditions $x_{1}(0)=1\mu{\rm m}$ , and $x_{2}(0)=-1\mu{\rm m}$ (corresponding to the second eigenvector V⁽²⁾, we get the plot shown in fig. 11.45. From this plot, we find that the time period of one oscillation is approximately $1.57\times 10^{-4}$ seconds, which gives a frequency of $\lambda_{n}=2\pi/T=4\times 10^{4}\,\mathrm{rad}/\mathrm{s}$ . Thus, the results of the numerical solution match the results obtained from the eigenvalue analysis.

Filename:bikefork1-ang-mom

Figure 11.45: Motions of mass A and mass B, $x_{1}(t)$ and $x_{2}(t)$ in the second normal mode. In this mode, each mass moves opposite to the other (out of phase) but both oscillate with the same frequency, $\lambda_{n}^{(2)}$ .

Answer:

Problems for 11.3 Normal Modes

11.3.1 A two degree of freedom mass-spring system, made up of two unequal masses $m_{1}$ and $m_{2}$ and three springs with unequal stiffnesses $k_{1}$ , $k_{2}$ and $k_{3}$ , is shown in the figure. All three springs are relaxed in the configuration shown. Neglect friction.

(a)

Derive the equations of motion for the two masses.
(b)

Does each mass undergo simple harmonic motion? Answer: If we start off by assuming that each mass undergoes simple harmonic motion at the same frequency but different amplitudes, we will find that this two-degree-of-freedom system has two natural frequencies. Associated with each natural frequency is a fixed ratio between the amplitudes of each mass. Each mass will undergo simple harmonic motion at one of the two natural frequencies only if the initial displacements of the masses are in the fixed ratio associated with that frequency.

Filename:bikefork-ang-mom

Figure 11.46

11.3.2 Normal Modes. Three equal springs ( $k$ ) hold two equal masses ( $m$ ) in place. There is no friction. $x_{1}$ and $x_{2}$ are the displacements of the masses from their equilibrium positions.

(a)

How many independent normal modes of vibration are there for this system? Answer: Two normal modes.
(b)

Assume the system is in a normal mode of vibration and it is observed that $x_{1}=A\sin(ct)+B\cos(ct)$ where $A$ , $B$ , and $c$ are constants. What is $x_{2}(t)$ ? (The answer is not unique. You may express your answer in terms of any of $A$ , $B$ , $c$ , $m$ and $k$ . ) Answer: $x_{2}=const*x_{1}=const*(A\sin(ct)+B\cos(ct))$ , where $const=\pm 1$ .
(c)

Find all of the frequencies of normal-mode-vibration for this system in terms of $m$ and $k$ . Answer: $\omega_{1}=\sqrt{\frac{3k}{m}}$ , $\omega_{2}=\sqrt{\frac{k}{m}}$ .

Filename:summer95f-5-a

Figure 11.47

11.3.3 $x_{1}(t)$ and $x_{2}(t)$ are measured positions on two points of a vibrating structure. $x_{1}(t)$ is shown. Some candidates for $x_{2}(t)$ are shown. Which of the $x_{2}(t)$ could possibly be associated with a normal mode vibration of the structure? Answer “could” or “could not” next to each choice and briefly explain your answer (If a curve looks like it is meant to be a sine/cosine curve, it is.)

Filename:pfigure4-2-rp10

Figure 11.48

More-Involved Problems

11.3.4 Two masses are connected to fixed supports and each other with the two springs and dashpot shown. The displacements $x_{1}$ and $x_{2}$ are defined so that $x_{1}=x_{2}=0$ when both springs are unstretched.

For the special case that $C=0$ and $F_{0}=0$ clearly define two different sets of initial conditions that lead to normal mode vibrations of this system.

Filename:pfigure-blue-125-2

Figure 11.49:

11.3.5 As in problem 10.62, a system of three masses, four springs, and one damper are connected as shown. Assume that all the springs are relaxed when $x_{A}=x_{B}=x_{D}=0$ .

(a)

In the special case when $k_{1}=k_{2}=k_{3}=k_{4}=k$ , $c_{1}=0$ , and $m_{A}=m_{B}=m_{D}=m$ , find a normal mode of vibration. Define it in any clear way and explain or show why it is a normal mode in any clear way. Answer: One normal mode: $[1,0,0]$ .
(b)

In the same special case as in (a) above, find another normal mode of vibration. Answer: The other two normal modes: $[0,1,\frac{1\pm\sqrt{17}}{4}]$ .

Filename:pfigure-blue-68-1

Figure 11.50:

11.3.6 As in problem 10.62, a system of three masses, four springs, and one damper are connected as shown. In the special case when $c_{1}=0$ , find the normal modes of vibration.

Filename:pfigure-blue-58-1

Figure 11.51:

11.3.7 Normal modes. All three masses have $m=1\,\mathrm{kg}$ and all 6 springs are $k=1\,{\mbox{N}}/\,\mathrm{m}$ . The system is at rest when $x_{1}=x_{2}=x_{3}=0$ .

(a)

Find as many different initial conditions as you can for which normal mode vibrations result. In each case, find the associated natural frequency. (we will call two initial conditions $[v]$ and $[w]$ different if there is no constant $c$ so that $[v_{1}\ v_{2}\ v_{3}]=c[w_{1}\ w_{2}\ w_{3}]$ . Assume the initial velocities are zero.)
(b)

For the initial condition $[x_{0}]=[\ 0.1\,\mathrm{m}\ \ 0\ \ 0\ ]$ , $[\dot{x}_{0}]=[\ 0\ \ 2\,\mathrm{m}/\mathrm{s}\ \ 0\ ]$ what is the initial (immediately after the start) acceleration of mass 2?

Filename:pfigure-blue-157-1

Figure 11.52:

11.3.8 For the three-mass system shown, assume $x_{1}=x_{2}=x_{3}=0$ when all the springs are fully relaxed. One of the normal modes is described with the initial condition $({x_{1}}_{0},x_{2},x_{3})=(1,0,-1)$ .

(a)

What is the angular frequency $\omega$ for this mode? Answer in terms of $L, m, k$ , and $g$ . (Hint: Note that in this mode of vibration the middle mass does not move.) Answer: $\omega=\sqrt{\frac{2k}{m}}$ .
(b)

Make a neat plot of $x_{3}$ versus $x_{1}$ for one cycle of vibration with this mode.
(c)

Make a single plot with both $x_{1}$ and $x_{3}$ vs $t$ for one cycle of this mode.

Filename:summer95f-5

Figure 11.53:

	$\displaystyle x(t)\ =$	$\displaystyle\phantom{+}Ae^{\left(-\frac{c}{2m}+\sqrt{(\frac{c}{2m})^{2}-\frac% {k}{m}}\right)t}$
		$\displaystyle+Be^{\left(-\frac{c}{2m}-\sqrt{(\frac{c}{2m})^{2}-\frac{k}{m}}% \right)t}.$		(11.6)

$\displaystyle\sum F$	$\displaystyle=$	$\displaystyle ma_{y}$
$\displaystyle\hbox to36.135pt{or\hfil}-k_{1}y-k_{2}y$	$\displaystyle=$	$\displaystyle m\ddot{y}$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{y}+\left(\frac{k_{1}+k_{2}}{m}% \right)y$	$\displaystyle=$	$\displaystyle 0.$

	$\displaystyle k_{2}y_{2}-k_{1}y_{1}$	$\displaystyle=$	$\displaystyle\underbrace{m}_{0}\;\;a=0$
	$\displaystyle y_{1}$	$\displaystyle=$	$\displaystyle\frac{k_{2}}{k_{1}}y_{2}$		(11.13)

$\displaystyle-k_{2}y_{2}=ma_{y}$	$\displaystyle=$	$\displaystyle m\ddot{y}$
$\displaystyle\hbox to36.135pt{or\hfil}m\ddot{y}+k_{2}\overbrace{\frac{k_{1}}{k% _{1}+k_{2}}y}^{y_{2}}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{y}+\frac{k_{1}k_{2}}{m(k_{1}+k_{2}% )}y$	$\displaystyle=$	$\displaystyle 0.$	(11.14)

	$\displaystyle\underbrace{-kx-c\dot{x}}_{\sum F_{x}}=m\underbrace{\ddot{x}}_{a_% {x}}$
	$\displaystyle\mbox{or}\quad\ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x=0$		(11.16)

$\displaystyle c^{2}$	$\displaystyle=$	$\displaystyle 2500(\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m})^{2}$
$\displaystyle\mbox{and}\quad 4mk$	$\displaystyle=$	$\displaystyle 4\mathord{\cdot}10\,\mathrm{kg}\mathord{\cdot}1000\,{\mbox{N}}/% \,\mathrm{m}=4000(\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m})^{2}.$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad c^{2}$	$\displaystyle<$	$\displaystyle 4mk.$

$\displaystyle x(0)$	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{m}=A.$
$\displaystyle\dot{x}(0)$	$\displaystyle=$	$\displaystyle 0=-\frac{c}{2m}\cdot A+\lambda_{D}\mathord{\cdot}B$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad B$	$\displaystyle=$	$\displaystyle\frac{A\,c}{2m\lambda_{D}}=\frac{(0.5\,\mathrm{m})\mathord{\cdot}% (50\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}/\,\mathrm{m})}{2\mathord{\cdot}(10% \,\mathrm{kg})\mathord{\cdot}(9.682\,\mathrm{rad}/\mathrm{s})}=0.13\,\mathrm{m}.$

$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle Ae^{-\xi\cdot\lambda\cdot\frac{\pi}{2\lambda_{D}}}$
	$\displaystyle=$	$\displaystyle Ae^{-\frac{\pi}{2}\cdot\frac{\xi}{\sqrt{1-\xi^{2}}}}$
	$\displaystyle=$	$\displaystyle(0.3\,\mathrm{m})\cdot e^{-0.0314}$
	$\displaystyle=$	$\displaystyle 0.29\,\mathrm{m}.$

	$\displaystyle\frac{E_{n+1}}{E_{n}}$	$\displaystyle=$	$\displaystyle\frac{\frac{1}{2}k(Ae^{-\xi\lambda(t_{n}+T_{D}})^{2}}{\frac{1}{2}% k(Ae^{-\xi\lambda t_{n}})^{2}}$
		$\displaystyle=$	$\displaystyle e^{-2\xi\lambda T_{D}}.$

$\displaystyle{E_{n+1}}$	$\displaystyle=$	$\displaystyle{E_{n}}\,e^{-2\xi\cancel{\lambda}\cdot\frac{2\pi}{\cancel{\lambda% }\sqrt{1-\xi^{2}}}}$
	$\displaystyle=$	$\displaystyle e^{-4\pi\frac{\xi}{\sqrt{1-\xi^{2}}}}\approx e^{-4\pi\xi}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad E_{n+1}$	$\displaystyle=$	$\displaystyle e^{-4\pi\xi}\,E_{n}.$

$\displaystyle E_{n}$	$\displaystyle=$	$\displaystyle e^{-4\pi\xi}\mathord{\cdot}E_{n-1}$
	$\displaystyle=$	$\displaystyle e^{-4\pi\xi}\mathord{\cdot}(e^{-4\pi\xi}\cdot E_{n-2})$
	$\displaystyle=$	$\displaystyle(e^{-4\pi\xi})^{3}\mathord{\cdot}E_{n-3}$
	$\displaystyle\vdots$
	$\displaystyle=$	$\displaystyle(e^{-4\pi\xi})^{n}\mathord{\cdot}E_{0}.$

$\displaystyle\Delta E_{5}$	$\displaystyle=$	$\displaystyle\frac{E_{0}-E_{5}}{E_{0}}\times 100$
	$\displaystyle=$	$\displaystyle\left(1-e^{-4\pi\xi\;\mathord{\cdot}\;5}\right)\times 100$
	$\displaystyle=$	$\displaystyle 71.5\%.$

	$\displaystyle F-kx$	$\displaystyle=$	$\displaystyle m\ddot{x}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\ddot{x}+kx$	$\displaystyle=$	$\displaystyle F$		(11.38)

	$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle(0.005\,\mathrm{m})\cos(44.72\,\mathrm{rad}/\mathrm{s}\cdot t)+0.% 005\,\mathrm{m},$
	$\displaystyle\hbox to36.135pt{and\hfil}\dot{x}(t)$	$\displaystyle=$	$\displaystyle-(0.22\,\mathrm{m}/\mathrm{s})\sin(44.72\,\mathrm{rad}/\mathrm{s}% \cdot t).$

	$\displaystyle X$	$\displaystyle=$	$\displaystyle\frac{F_{0}/k}{\sqrt{(2\xi r)^{2}+(1-r^{2})^{2}}}\approx\frac{F_{% 0}/k}{1}=\frac{F_{0}}{k}$
	$\displaystyle\phi$	$\displaystyle=$	$\displaystyle\tan^{-1}(2\xi r)\approx\tan^{-1}0=0$

$\displaystyle\mbox{Mass A:}\qquad m\ddot{x}_{1}$	$\displaystyle=$	$\displaystyle k_{2}(x_{2}-x_{1})-2k_{1}x_{1}$
	$\displaystyle=$	$\displaystyle-(2k_{1}+k_{2})x_{1}+k_{2}x_{2}$
$\displaystyle\mbox{Mass B:}\qquad m\ddot{x}_{2}$	$\displaystyle=$	$\displaystyle-k_{2}(x_{2}-x_{1})-2k_{1}x_{2}$
	$\displaystyle=$	$\displaystyle k_{2}x_{1}-(2k_{1}+k_{2})x_{2}.$

	$\displaystyle\lambda^{(1)}=1.449\times 10^{9},$		$\displaystyle{\rm\bf V}^{(1)}=\left(\begin{array}[]{c}1\\ 1\end{array}\right);$
	$\displaystyle\hbox to36.135pt{\hfil and\hfil}\lambda^{(2)}=1.623\times 10^{9},$		$\displaystyle{\rm\bf V}^{(2)}=\left(\begin{array}[]{r}1\\ -1\end{array}\right).$

	$\displaystyle\lambda_{n}^{(1)}=3.807\times 10^{4}\,\mathrm{rad}/\mathrm{s}$	$\Rightarrow$	$\displaystyle f_{n}^{(1)}=\lambda_{n}^{(1)}/2\pi=6.06\;{\rm kHz},$
	$\displaystyle\hbox to36.135pt{and\hfil}\lambda_{n}^{(2)}=4.029\times 10^{4}\,% \mathrm{rad}/\mathrm{s}$	$\Rightarrow$	$\displaystyle f_{n}^{(2)}=\lambda_{n}^{(2)}/2\pi=6.41\;{\rm kHz}.$

$\displaystyle\dot{x}_{1}$	$\displaystyle=$	$\displaystyle u_{1}$
$\displaystyle\dot{u}_{1}$	$\displaystyle=$	$\displaystyle-\frac{(2k_{1}+k_{2})}{m}x_{1}+\frac{k_{2}}{m}x_{2}$
$\displaystyle\dot{x}_{2}$	$\displaystyle=$	$\displaystyle u_{2}$
$\displaystyle\dot{u}_{2}$	$\displaystyle=$	$\displaystyle\frac{k_{2}}{m}x_{2}-\frac{(2k_{1}+k_{2})}{m}x_{2}.$