Solution Manual Engineering Vibration 5th Edition by Inman – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Problems and Solutions Section 1.1 (1.1 through 1.27)

1.1 A spring-mass system has a mass of 100 kg and a stiffness of 10,000 N/m. What is

its period of oscillation?

Solution:

1.2 k

10,000 N/m

100 kg

= 10 rad/s Þ T=

10= 0.62 s

The mass of a passenger car is about 2500 kg and has a stiffness of 161,255 N/m.

Compare the frequency and period of the car empty to that if 181 kg of passengers

are added to the car.

1.3 Solution: The frequency and period of the empty car are:

161,255 N/m

2500 kg

= 8.031 rad/s Þ T=

8.031= 0.81 s

With 181 kg of passengers the frequency and period become

161,255 N/m

2681 kg

= 7.555 rad/s Þ T=

7.555= 0.78 s

Thus in this case the difference in period of oscillation with and without the

passengers is an imperceptible 3 hundredths of a second. For lighter cars this

difference could become perceptible.

Consider a simple pendulum (see Example 1.1.1) and compute the magnitude of the

restoring force if the mass of the pendulum is 2 kg and the length of the pendulum

is 0.5 m. Assume the pendulum is at the surface of the earth at sea level.

Solution: From example 1.1.1, the restoring force of the pendulum is mgl sinq,

which has maximum value

1.4 mgl= 2×9.81×0.5 kg×m ×m

sec2

= 9.81 N ×m

Compute the period of oscillation of a pendulum of length 1 m at the North Pole

where the acceleration due to gravity is measured to be 9.832 m/s2

relationships:

The spring of Figure 1.2, repeated here as Figure P1.3, is loaded with mass of 15 kg

and the corresponding (static) displacement is 0.01 m. Calculate the spring’s

stiffness.

Figure P1.3

Solution:

Free-body diagram:

From the free-body diagram and static

equilibrium:

kx = mg k= mg / x

15 × 9.81

0.01

(g= 9.81m / s

2 )

= 14715 N/m

(static) displacement is recorded below. Plot the data and calculate the spring’s

stiffness. Note that the data contain some error. Also calculate the standard

deviation.

m(kg) 10 11 12 13 14 15 16

x(m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82

Solution:

Free-body diagram:

From the free-body diagram and static

equilibrium:

kx = mg k= mg / x

(g= 9.81m / s

2 )

Ski

= 86.164

The sample standard deviation in

computed stiffness is:

i=1

s =

(ki– m)2

n– 1= 0.164

0 1 2

Computation of slope from mg/x

m(kg) x(m) k(N/m)

10 1.14 86.05

11 1.25 86.33

12 1.37 85.93

13 1.48 86.17

14 1.59 86.38

15 1.71 86.05

16 1.82 86.24

Consider the pendulum of Example 1.1.1 and compute the amplitude of the

restoring force if the mass of the pendulum is 2 kg and the length of the pendulum

is 0.5 m if the pendulum is at the surface of the moon.

Solution: From example 1.1.1, the restoring force of the pendulum is mgl sinq,

which has maximum value

9.81

1.5 mgl= 2×

×0.5 kg×m ×m

sec2

= 1.635 N ×m

Consider the pendulum of Example 1.1.1 and compute the angular natural

frequency (radians per second) of vibration for the linearized system if the mass of

the pendulum is 2 kg and the length of the pendulum is 0.5 m if the pendulum is at

the surface of the earth. What is the period of oscillation in seconds?

Solution: The natural frequency and period are:

the case with n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s.

Solution:

Given:

mx + kx = 0 (1)

Assume: x(t)= ae

rt . Then: x = are

rt and x = ar

rt . Substitute into equation (1) to

get:

mar

rt + kae

2 + k= 0

r = ±

= 0

Thus there are two solutions:

è ç

x1 = c1e

ø ÷ t

è ç

, and x2 = c2e–

ø ÷ t

where wn

= 2 rad/s

The sum of x1 and x2 is also a solution so that the total solution is:

x = x1 + x2 = c1e

2it + c2e–2it

Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/s

x 0

( )= c1 + c2 = x0 = 1Þ c2 = 1– c1, and v 0

( )= x 0

( )= 2ic1– 2ic2 = v0 = 5 mm/s

–2c1 + 2c2 = 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns):

–2c1 + 2– 2c1 = 5 iÞ c1 =

2–

i, and c2 =

x =

è ç

2–

ø ÷ e

2it +

è ç

ø ÷ e–2it

Using the Euler formula to evaluate the exponential terms yields:

x =

è ç

2–

( )+

ø ÷ cos2t + i sin2t

è ç

( )

ø ÷ cos2t– i sin2t

Þ x(t) = cos2t +

sin2t=

sin 2t + 0.7297

( )

Using Mathcad the plot is:

x t cos . 2 t .

sin . 2 t

x t

0 5 10

1.10 Solve mx + kx = 0 for k = 4 N/m, m = 1 kg, x0 = 1 mm, and v0 = 0. Plot the

solution.

Solution: Here v0 = 0. æ

è ç

= 2 rad/s

ø ÷ . Calculating the initial conditions:

x 0

( )= c1 + c2 = x0 = 1Þ c2 = 1– c1

v 0

( )= x 0

( )= 2ic1– 2ic2 = v0 = 0Þ c2 = c1

c2 = c1 = 0.5

x t

( )=

2it +

e–2it

cos2t + i sin2t

( )+

x(t)= cos (2t )

cos2t– i sin2t

( )

x t cos . 2 t

x t

0 5 10

1.11 1.12 t

Alternately students may use equation (1.10) directly to get

x(t)=

22 (1)2 + 02

sin(2t + tan–1[2 ×1

0 ])

= 1sin(2t +

2 )= cos2t

The amplitude of vibration of a spring-mass system is measured to be 1 mm. The

phase shift from t = 0 is measured to be 2 rad and the frequency is found to be

5 rad/s. Calculate the initial conditions that caused this vibration to occur. Assume

the response is of the form x(t)= Asin(wnt +f ).

Solution:

Given: A= 1mm,f= 2rad, w = 5rad/s. For an undamped system:

x t

( )= Asin wnt +f

( )= 1sin 5t + 2

( ) and

v t

( )= x t

( )= Awn cos wnt +f

( )= 5cos 5t + 2

( )

Setting t = 0 in these expressions yields:

x(0) = 1sin(2) = 0.9093 mm

v(0) = 5 cos(2) = – 2.081 mm/s

Determine the stiffness of a single-degree-freedom, spring-mass system with a

mass of 100 kg such that the natural frequency is 10 Hz.

frequency:

1.13 10 Hz = 10 cycle

sec

2prad

cycle

= 20p rad / sec

mÞ k= mw

= 100kg(20p)2 1

sec2

= 394,784 N/m

Find the equation of motion for the system of Figure P1.11, and find the natural

frequency. In particular, using static equilibrium along with Newton’s law,

determine what effect gravity has on the equation of motion and the system’s

natural frequency. Assume the block slides without friction.

Figure P1.11

Solution:

Choosing a coordinate system along the plane with positive down the plane, the

free-body diagram of the system for the static case is given and (a) and for the

dynamic case in (b):

by the angle  as indicated. From the static equilibrium:–kxs + mgsinq= 0.

Summing forces in (b) yields:

F iå

= mx(t)Þ mx(t)= –k(x + xs ) + mg sinq

Þ mx(t) + kx = –kxs + mg sinq= 0

Þ mx(t) + kx = 0

Þ wn

rad/s

An undamped system vibrates with a frequency of 10 Hz and amplitude 1 mm.

Calculate the maximum amplitude of the system’s velocity and acceleration.

Solution:

Given: First convert Hertz to rad/s: wn

= 2pfn

= 2p 10 ( )= 20p rad/s.We also have

that A= 1 mm.

For an undamped system:

x t

( )= Asin wnt +f

( )

and differentiating yields the velocity: v t

( )= A wn cos wnt +f

( ). Realizing that both

the sin and cos functions have maximum values of 1 yields:

vmax

= Awn

= 1 20p

( )= 62.8 mm / s

Likewise for the acceleration: a t

( )= –Awn

2 sin wnt +f

( )

amax

= Awn

= 1 20p

( )2

= 3948 mm / s2

Show by calculation that A sin (nt + ) can be represented as A1sin nt + A2 cosnt

and calculate A1 and A2 in terms of A and .

Solution:

This trig identity is useful: sin a + b

( )= sina cosb + cosasinb

Given: Asin wnt +f

( )= Asin wnt

( )cos f ( )+ Acos wnt

( )sin f ( )

where A 1

= A 1 sinw

t + A2 cosw

= Acosf and A2

= Asinf

t + A2 cosw

calculate the values of A1and A2 in terms of the initial conditions x0 and v0.

Solution:

Using the solution of equation (1.2) in the form

x t

( )= A 1 sinw

t + A2 cosw

and differentiate to get:

x(t)= w

A 1 cos(w

t)-w

A2 sin(w

Now substitute the initial conditions into these expressions for the position and

velocity to get:

= x(0)= A 1 sin(0) + A2 cos(0)= A2

= x(0)= w

A 1 cos(0)-w

A2 sin(0)

= w

A 1(1)-w

A2 (0)= w

A 1

Solving for A1 and A2 yields:

A 1

, and A2

= x0

Thus x(t)=

sinwnt + x0 coswnt

1.17 Using the drawing in Figure 1.7, verify that equation (1.10) satisfies the initial

velocity condition.

Solution: Following the lead given in Example 1.1.2, write down the general

expression of the velocity by differentiating equation (1.10):

x(t)= Asin(wnt +f )Þ x(t)= Awn cos(wnt +f )

Þ v(0)= Awn cos(wn 0 +f )= Awn cos(f )

Figure 1.7

A= x0

2 +

è ç

ø ÷

, cosf=

2 +

è ç

ø ÷

Substitution of these values into the expression for v(0) yields

v(0)= Awn cosf= x0

2 +

è ç

ø ÷

(wn )

2 +

è ç

= v0

ø ÷

verifying the agreement between the figure and the initial velocity condition.

A 0.5 kg mass is attached to a linear spring of stiffness 0.1 N/m. a) Determine the

natural frequency of the system in hertz. b) Repeat this calculation for a mass of

50 kg and a stiffness of 10 N/m. Compare your result to that of part a.

Solution: From the definition of frequency and equation (1.12)

( ) wn

0.5

0.1= 0.447 rad/s

2.236

= 0.071 Hz

10= 0.447rad/s, fn

= 0.071 Hz

Part (b) is the same as part (a) thus very different systems can have same natural

frequencies.

Newton’s law, ma = –kx, in differential form using adx = vdv and integrating twice.

Solution: Substitute a = vdv/dx into the equation of motion ma = –kx, to get

mvdv = –kxdx. Integrating yields:

2 x

2=-wn

+ c

, where c is a constant

or v

=-wn

2 + c

2 Þ

v =

dt=-wn

2 + c

2 Þ

dt=

2 , write u = wn x to get:

-wn

2 + c

t– 0=

ò=

– u

sin–1 u

è ç ö

ø ÷ + c2

Here c2 is a second constant of integration that is convenient to write as

c2 = –/n. Rearranging yields

wnt +f= sin–1 wn x

è ç ö

ø ÷ Þ

wn x

= sin(wnt +f ) Þ

x(t)= A sin(wnt +f ), A=

in agreement with equation (1.19).

Determine the natural frequency of the two systems illustrated.

k1 k2

(a) (b)

Figure P1.18

Solution:

(a) Summing forces from the free-body diagram in the x direction yields:

Examining the coefficient of x

yields:

Free-body diagram for part a

–k2x

–k1x

k1 + k2

mx =-k1x– k2 x Þ

mx + k1x + k2 x = 0

mx + x k1 + k2

( )= 0, dividing by m yields:

k1 + k2

x +

è ç ö

ø ÷ x = 0

(b) Summing forces from the free-body diagram in the x direction yields:

–k1x

–k3x

–k2x

Free-body diagram for part b

mx = –k1x– k2 x– k3x,Þ

mx + k1x + k2 x + k3x = 0Þ

mx + (k1 + k2 + k3 )x = 0Þ x + (k1 + k2 + k3 )

x = 0

Þ wn

k1 + k2 + k3

1.21* Plot the solution given by equation (1.11) for the case k = 1000 N/m and m = 10 kg

for two complete periods for each of the following sets of initial conditions: a) x0 =

0 m, v0 = 1 m/s, b) x0 = 0.01 m, v0 = 0 m/s, and c) x0 = 0.01 m, v0 = 1 m/s.

a) all units in m, kg, s

m 10

k 1000

x0 0.0

. 2 p

v0 1

. 2 p

f atan

wn x0

x t . A sin .

wn t f

parts b and c are plotted in the above by simply changing the initial conditions as

appropriate

A .

x02 wn

v02

0.2

0.1

x t

xb t

xc t

0 0.5 1 1.5

0.1

0.2

Figure P1.21. Ignore the mass of pendulum’s rod and derive the equation of motion.

Then following the procedure used in Example 1.1.1, linearize the equation of

motion and compute the formula for the natural frequency. Assume that the rotation

is small enough so that the spring only deflects horizontally.

Figure P1.21

Solution: Consider the free body diagram of the mass displaced from equilibrium:

There are two forces acting on the system to consider, if we take moments about

point O (then we can ignore any forces at O). This yields

MOå

= JOa Þ m

2q= –mg sinq- k sinq· cosq

Þ m

2q+ mg sinq+ k 2 sinqcosq= 0

mg + k

q (t) +

è ç ö

ø ÷ q (t)= 0

. Then the

Thus the natural frequency is

mg + k

rad/s

A pendulum has length of 250 mm. What is the system’s natural frequency in

Hertz?

Solution:

Given: l =250 mm

Assumptions: small angle approximation of sin

From Window 1.1, the equation of motion for the pendulum is as

follows: IOq+ mg q= 0, where IO = ml2

Þq+ g

q= 0

The coefficient of  yields the natural frequency as:

   

9.8 m/s 6.26 rad/s

0.25 mn

f 

 



0.996 Hz

The pendulum in Example 1.1.1 is required to oscillate once every second. What

length should it be?

Solution:

Given: f = 1 Hz (one cycle per second)

   

f l

g l m

   

9.81 0.248

2 2

(2 ) 4

f  

length 0.5 m, has an initial position of  0) = 0, what is the maximum value of the

initial angular velocity that can be given to the pendulum with out violating this

small angle approximation? (be sure to work in radians)

Solution: From Window 1.1, the linear equation of the pendulum is

( ) ( ) 0 g t t     

For zero initial position, the solution is given in equation (1.10) by

q(t)=

sin( gt)Þ q £

since sin is always less then one. Thus if we need  < 10°= 0.175 rad, then we need

to solve:

5 . 0

81 . 9

175 . 0

for v0 which yields:

1.26 v0 < 0.773 rad/s.

A machine, modeled as a simple spring-mass system, oscillates in simple harmonic

motion. Its acceleration is measured to have an amplitude of 10,000 mm/s2 with a

frequency of 8 Hz. Compute the maximum displacement the machine undergoes

during this oscillation.

Solution: the equations of motion for position and acceleration are:

x = Asin(w

t +f ) and x = –A w

2 sin(w

t +f )

Since the sin is max at 1, the maximum acceleration is

A w

= 10,000 mm/s2

= 2pf= 2p(8)= 16p rad/s

10,000

(16p)2

= 3.96 mm

Derive the relationships given in Window 1.4 for the constants a1 and a2 used in the

exponential form of the solution in terms of the constants A1 and A2 used in sum of

sine and cosine form of the solution. Use the Euler relationships for sine and cosine

in terms of exponentials as given following equation (1.18).

Solution: Let  = t for ease of notation. Then:

2sinqj= eqj

– eqj and 2cosq= eqj + eqj

Þ A 1 sinq= A 1

– e

and A2 cos = A2

2 j

qj + e

Adding these to in order to form x(t) yields:

x(t)= A 1

2 j

– A 1

e-qj

2 j

+ A2

e-qj

Þ x(t)= –A 1

eqj

2 j + A 1

Þ x(t)= (A2– A 1 j) e

e-qj

eqj

2 j + A2

+ (A 1 j + A2 ) e-qj

+ A2

e-qj

Comparing this last expression to x(t)= a1e

qj + a2e-qj yields:

A2– A 1 j

and a2

A2 + A 1 j

For a pendulum at the earth’s surface, plot the frequency versus the length of the

pendulum for values of the length between 0 and 4 meters.

Solution: Using the formula for frequency of a pendulum, typing in MATLAB

>> L=0:1/100:4;

>> x=sqrt((1./L)*9.81); % use the dot to perform element by element division

>> plot(L,x)

>> xlabel(‘lenght in meters’)

>> ylabel(‘frequency in rad/s’)

Plot how the frequency changes in a 2500 kg car as the possible stiffness values

range from 120,000 N/m to 170,000 N/m. Express your answer in Hz.

Solution: Typing in MATLAB’s command window

>> m=2500;

>> K=120000:50000/500:170000;

>> omega=sqrt((1/m).*K);

>> w=omega/(2*pi);

>> plot(K,w)

>> xlabel(‘stiffness in N/m’)

>> ylabel(‘frequency in Hz’)

Consider the system of Problem 1.13 and Figure P1.13. Suppose the surface of the

plane provides Coulomb friction and determine the equation for vibration.

Solution: Choosing a coordinate system along the plane with positive down the

plane, the free-body diagram of the system for the static case is given and (a) and

for the dynamic case in (b):

by the angle  as indicated. From the static equilibrium:–kxs + mgsinq= 0.

Summing forces in (b) yields:

Problems and Solutions Section 1.2 (Numbers 1.31 through 1.48)

1.31 The acceleration of a machine part modeled as a spring mass system is measured

and recorded in Figure P 1.31. Compute the amplitude of the displacement of the

mass.

Figure P1.31

Solution: From Window 1.3 the maximum amplitude of the acceleration versus

time plot is just w

2 A where A is the maximum amplitude of the displacement and

the quantity to be determined here. Looking at P1.31, not that the plot repeats

itself twice after 2.5 s so that T = 2.5/2 = 1.25 s. Also the plot has 1 m/s2 as its

maximum value. Thus w

2 A= 1 and

2 m/s2

1 m/s2

è ç

ø ÷

è ç

ø ÷

m =

è ç

1.25

ø ÷

m = 0.0396 m

Solution: First change the mass given in kg into slugs using 1 lb-sec2/

ft = 14.5939 kg. So

m = 49.2 ´10–3 kg ´

1 lb×sec2 / ft

14.5939 kg

= 3.37127 ´10–3 lb×sec2 / ft

Using the conversions: 1 N = 0.224808 lb and 1 m = 3.280839 ft, the stiffness

becomes:

857.8 N

0.224808 lb

1 m

3.280839 ft= 58.78 lb

Thus the frequency becomes

58.78 lb/ft

3.37127 (lb/ft)×sec2

= 132.05 rad/s

which agrees with the example as it should.

The period is of course the same. To compute the maximum amplitude change

mm to in using 1 millimeter = 0.039 370 078 74 inch. So x0 = 10 mm = 0.39 in

and the max amplitude becomes A = 0.39 in and the max acceleration becomes

Since the velocity is zero the phase is 90° and the solution is

x(t)= 0.39cos(132t) in

Which is identical to the solution in mm.

Referring to Example 1.2.2, determine the length in feet by using the formula for

the period as done in the Example.

Solution: From the example the formula for the length of a pendulum is

gT 2

4p2

(32.174 ft/s2 ) 3 s

( )2

= 7.335 ft

4p2

As a check 2.237 m = 2.237 m x (3.280 ft/m) = 7.339 ft a little difference in round

off.

Calculate the moon’s acceleration due to gravity in ft/s2

Solution:

= g / 6= (32.174 ft/s2 ) / 6= 5.362 ft/s2

mm/s2 and measured displacement amplitude of 2 mm. Calculate the systems

natural frequency.

Solution: The amplitude of displacement is A = 2mm, and that of acceleration is

2 A= 8Þw

= 4Þw

= 2 rad/s

1.36 A spring-mass system has measured period of 5 seconds and a known mass of

20 kg. Calculate the spring stiffness.

Solution: Using the basic formulas for period and frequency:

= 5 s Þ k=

è ç

ø ÷

´ m =

è ç

ø ÷

20 kg

= 31.583 N/m

1.37* Plot the solution of a linear, spring and mass system with frequency n =1 rad/s,

x0 = 2 mm and v0 = 2 mm/s, for at least two periods.

Solution: From the formula in Window 1.2, the plot can be formed by computing:

2 + v0

= 2.8284 mm, f= tan–1 wn x0

è ç

ø ÷= 0.7854 rad/s

x(t)= Asin(wnt +f )= 2.2sin(t + 0.7854)

The period is 2/n =2, so the plot needs to run to 4.

The solution in Matlab is

>> wn=1;x0=2;v0=2;

>> A=sqrt(wn^2*x0^2+v0^2)/wn;

>> p=atan2(wn*x0,v0);

>> figure

>> t=(0:0.01:4*pi);

>> x=A*sin(wn*t+pi/2);

>> plot(t,x)

>> xlabel(‘time in seconds’)

>> ylabel(‘displacement in mm’

>> A

>>A = 2.8284

p = 0.7854

Compute the natural frequency and plot the solution of a spring-mass system with

mass of 1 kg and stiffness of 4 N/m, and initial conditions of x0 = 2 mm and

v0 = 0 mm/s, for at least two periods.

Solution: Working entirely in the command window of Matlab, and using the

units of mm yields:

>> m=1;k=4;x0=2;v0=0;

>> wn=sqrt(k/m)

wn = 2

>> A=(1/wn)*sqrt(wn^2*x0^2+v0^2)

A = 2

>> figure

>> t=[0:0.01:6];

>> x=A*sin(wn*t+pi/2);

>> plot(t,x)

>> xlabel(‘time in seconds’)

>> ylabel(‘displacement in mm’)

spring constant such that the resulting natural frequency has a specified value.

Suppose that the mass of a system is 4 kg and the stiffness is 100 N/m. How much

must the spring stiffness be changed in order to increase the natural frequency by

10%?

Solution: Given m = 4 kg and k = 100 N/m the natural frequency is

100

4= 5 rad/s

Increasing this value by 10% requires the new frequency to be 5 x 1.1 = 5.5 rad/s.

Solving for k given m and n yields:

5.5=

4Þ k= (5.5)2(4)=121 N/m

Thus the stiffness k must be increased by about 20%.

The pendulum in the Chicago Museum of Science and Industry has a length of

20 m and the acceleration due to gravity at that location is known to be 9.803

m/s2

. Calculate the period of this pendulum.

Solution: Following along through Example 1.2.2:

g / l=

9.803 / 20= 9.975 s

Calculate the RMS values of displacement, velocity and acceleration for the

undamped single degree of freedom system of equation (1.19) with zero phase.

Solution: Calculate RMS values

Let

x t

( )= Asin wnt

˙ x t

( )= Awn coswn t

˙ ˙ x t

( )= –Awn

2 sinwn t

Mean Square Value: x

lim

T ® ¥

ò (t) dt

lim

T ® ¥

A2 sin2

ò dt=

wn t

lim

T ® ¥

ò ) dt=

T (1– cos 2wn t

x.. 2

lim

T® ¥

lim

T® ¥

Therefore,

ò dt=

cos

wn t

ò dt=

4 sin2

wn t

lim

T®¥

lim

T® ¥

A2wn

ò ) dt=

2 (1 + cos 2wn t

ò ) dt=

2 (1 + cos 2wn t

A2wn

xrms 2

rms = x. 2

Awn

x..

rms = x..

Awn

Calculate the coefficients A1 and A2 of the solution given in Window 1.4 and write

down the solution in that form for the case x0 =1 mm, vo = 1mm/s and the natural

frequency is 10 rad/s.

Solution:

= x(0)= A1 sin(0) + A2 cos(0)= A2

= 1 mm

v0 = v(0)= wn A1 cos(0)– A2wn sin(0)= 0.1 mm/s

Þ A1 = (0.1 mm/s) / 10 rad/sÞ A1 = 0.01 mm

Þ x(t)= 0.01sin(10t) +1cos(10t)

[ ] mm

A foot pedal mechanism for a machine is crudely modeled as a pendulum

connected to a spring as illustrated in Figure P1.43. The purpose of the spring is

to keep the pedal roughly vertical. Compute the spring stiffness needed to keep

the pendulum at 1° from the horizontal and then compute the corresponding

natural frequency. Assume that the angular deflections are small, such that the

spring deflection can be approximated by the arc length, that the pedal may be

treated as a point mass and that pendulum rod has negligible mass. The values in

the figure are m = 0.5 kg, g = 9.8 m/s2

, 1 = 0.2 m and 2 = 0.3 m.

Solution: You may want to note to your students, that many systems with springs are

often designed based on static deflections, to hold parts in specific positions as in this

case, and yet allow some motion. The free-body diagram for the system is given in

the figure.

For static equilibrium the sum of moments about point O yields ( 1 is the static

deflection):

M0å

= –

1 q 1 1

( )k + mg 2 = 0

Þ 1

2q 1k= mg 2

Þ k=

mg 2

2q 1

0.5×9.8×0.3

( )2 p

0.2

= 2106 N/m

(1)

Again take moments about point O to get the dynamic equation of motion:

MOå

= Jq= m 2

2q= –

2k(q+q 1) + mg 2 = –

2kq+ 1

2kq 1– mg 2q

m 2

2q+ 1

2 kq= 0

Þ q+ 1

è ç

m 2

ø ÷ q= 0

0.2

Þ wn

0.3

2106

0.5= 43.27 rad/s

An automobile is modeled as a 1000-kg mass supported by a spring of stiffness

k = 400,000 N/m. When it oscillates it does so with a maximum deflection of

10 cm. When loaded with passengers, the mass increases to as much as 1300 kg.

Calculate the change in frequency, velocity amplitude, and acceleration amplitude

if the maximum deflection remains 10 cm.

Solution:

Given: m1 = 1000 kg

m2 = 1300 kg

k = 400,000 N/m

xmax = A = 10 cm

000 , 400

n /

1 =

rad

1000

000 , 400

n /

2 =

54 . 17

rad

1300

w D

54 . 17–

46 . 2

rad /

Df=

=-2.46

= 0.392 Hz

v1 = A n1 = 10 cm  20 rad/s = 200 cm/s

v2 = A n2 = 10 cm  17.54 rad/s = 175.4 cm/s

 v = 175.4 – 200 = –24.6 cm/s

a1 = A n1

2 = 10 cm  (20 rad/s)2 = 4000 cm/s2

a2 = A n2

2 = 10 cm  (17.54 rad/s)2 = 3077 cm/s2

 a = 3077 – 4000 = –923 cm/s2

P1.45 to improve the car’s handling. (a) Compute the frequency of vibration of

the wheel assembly given that the torsional stiffness is 2000 N m/rad and the

wheel assembly has a mass of 38 kg. Take the distance x = 0.26 m.

(b) Sometimes owners put different wheels and tires on a car to enhance the

appearance or performance. Suppose a thinner tire is put on with a larger wheel

raising the mass to 45 kg. What effect does this have on the frequency?

1.46 Figure P1.45

Solution: (a) Ignoring the moment of inertial of the rod, and computing the

moment of inertia of the wheel as mx

2, the frequency of the shaft mass system is

2000 N × m

38 × kg (0.26 m)2

= 27.9 rad/s

(b) The same calculation with 45 kg will reduce the frequency to

2000 N × m

45 × kg (0.26 m)2

= 25.6 rad/s

This corresponds to about an 8% change in unsprung frequency and could

influence wheel hop etc. You could also ask students to examine the effect of

increasing x, as commonly done on some trucks to extend the wheels out for

appearance sake.

A machine oscillates in simple harmonic motion and appears to be well modeled

by an undamped single-degree-of-freedom oscillation. Its acceleration is

measured to have an amplitude of 10,000 mm/s2 at 7 Hz. What is the machine’s

maximum displacement?

Given: amax = 10,000 mm/s2 @ 7 Hz

The equations of motion for position and acceleration are:

x = Asin(wnt +f ) (1.3)

x = –Awn

2 sin(wnt +f ) (1.5)

The amplitude of acceleration is Aw 2

14 rad/s, from equation (1.12).

The machine’s displacement is A=

000 , 10

mm/s2 and n = 2f = 2(7) =

1.47 10,000

10,000

(14p)2

= 5.169 mm

A simple undamped spring-mass system is set into motion from rest by giving it

an initial velocity of 100 mm/s. It oscillates with a maximum amplitude of

15 mm. What is its natural frequency?

Solution:

Given: x0 = 0, v0 = 100 mm/s, A = 15 mm

From equation (1.9),

= or wn

100

15= 6.667, so that: n= 6.667 rad/s.

1.48 An automobile exhibits a vertical oscillating displacement of maximum amplitude

1 cm and a measured maximum acceleration of 2000 cm/s2. Assuming that the

automobile can be modeled as a single-degree-of-freedom system in the vertical

direction, calculate the natural frequency of the automobile.

Solution:

Given: A = 1 cm. From equation (1.15)

x w

2000

cm/s

Solving for n yields:

2000

1= 44.72 rad/s

Problems Section 1.3 (Numbers 1.49 through 1.72)

1.49 Consider a spring mass damper system, like the one in Figure 1.10, with the

following values: m =10 kg, c = 3 N/s and k = 1000 N/m. a) Is the system

+ x0wd

( )2

overdamped, underdamped or critically damped? b) Compute the solution if the

system is given initial conditions x0 = 0.01 m and v0 = 0.

Solution: a) Using equation 1.30 the damping ratio is

2 km

2 10×1000= 0.015 < 1

Thus the system is underdamped.

b) Using equations (1.38) the amplitude and phase can be calculated from the

initial conditions:

( )2

v0 +z wn x0

9.999 (0.015×10 ×0.01)2 + 0.01×9.999

( )2

= 0.01 m

f = tan–1 x0wd

v0 + zw

= tan–1 1– z2

= 1.556 rad

So the solution is Ae-z wnt sin wdt +f

( )= 0.01e–0.15t sin 9.999t +1.556

( ) m.

Note that for any system with v0 = 0 the phase is strictly a function of the damping

ratio. Also note that the code given below can be used to generate many problems

for homework or quizzes by just rearranging the numbers, always making sure to

keep the damping low enough to be underdamped. Typing the following in the

command window in Matlab:

>> m=10;c=3;k=1000;x0=0.01;v0=0.0;

>> wn=sqrt(k/m);

>> z=c/(2*sqrt(k*m));

>> wd=wn*sqrt(1-z^2);

>> A=sqrt(((v0+z*wn*x0)^2*(x0*wd)^2)/wd^2);p=atan2(wd*x0,

(v0+z*wn*x0));

>> t=[0:0.1:30];

>> x=A*exp(-z*wn*t).*sin(wn*t+p);

>> figure

>> plot(t,x)

>> ylabel(‘displacement in m’)

1.50 Consider a spring-mass-damper system with equation of motion given by

. Compute the damping ratio and determine if the system is

overdamped, underdamped or critically damped.

Solution: The parameter values are m = 1, k = 2 and c = 0.2. From equation (1.30)

the damping ratio is

2 km

0.2

2 2×1= 0.0707 <1

Hence the system is underdamped.

1.51 Consider the system 0

x for x0 = 1 mm, v0 = 0 mm/s. Is this system

+ x

overdamped, underdamped or critically damped? Compute the solution and

determine which root dominates as time goes on (that is, one root will die out

quickly and the other will persist.

Solution: From equation (1.30) the damping ratio is

2 km

2 1×1= 2 >1

Hence the system is overdamped.

Given 0

+ v

where 0

4 0

mm,

x = ae

Þ x = are

Þ x = ar

Substitute these into the equation of motion to get:

rt + 4are

rt + ae

= 0

2 + 4r + 1= 0Þ

r 1,2 = –2 ± 3

x = a1e–2 + 3

( )t

+ a2e– 2– 3

( )t

˙ x = – 2 + 3

( )a1e–2+ 3

( )t

+ – 2– 3

( )a2e–2– 3

( )t

Applying initial conditions yields,

x0 = a1 + a2Þ x0– a2 = a1 (1)

v0 = – 2 + 3

( )a1 + – 2– 3

( )a2 (2)

Substitute equation (1) into (2)

v0 = – 2 + 3

( )(x0– a2 ) + – 2– 3

( )a2

v0 = – 2 + 3

( )x0– 2 3 a2

Solve for a2

a2 =

–v0 + – 2 + 3

( )x0

2 3

Substituting the value of a2 into equation (1), and solving for a1 yields,

a1 =

v0 + 2 + 3

( )x0

2 3

\ x(t)=

v0 + 2 + 3

( )x0

2 3

( )t

–v0 + – 2 + 3

e–2+ 3

( )x0

2 3

( )t

e–2– 3

The response is dominated by the root: –2 + 3 as the other root dies off

very fast.

1.52 Compute the solution to 0

x for x0 = 0 mm, v0 = 1 mm/s and write

+ x

down the closed form expression for the response.

Solution:

The parameter values are m = 1, k = 2 and c = 2. From equation (1.30) the

damping ratio is

2 km

2 2×1= 0.707 <1

1= 2

Thus equations (1.36) and (1.38) can be used directly or one can follow the last

expression in Example 1.3.3:

x(t)= e-z wnt v0 +z wn x0

è ç

sinwd t + x0 coswd t

ø ÷

= e–

2t v0

è ç

sinwd t

ø ÷

The damped natural frequency is

wd = wn 1-z2

= 2 × 1–

2= 1

Thus the solution is

x(t)= e–t sint mm

Alternately use equations (1.36) and (1.38). The plot is similar to figure 1.12.

Derive the form of 1 and 2 given by equation (1.31) from equation (1.28) and

the definition of the damping ratio.

Solution:

Equation (1.28): km

–

c 4

, 1–

Rewrite, l 1,2 =-

è ç ö

2 m m

ø ÷ k

è ç

ø ÷ ±

2 m m

è ç

ø ÷

è ç ö

ø ÷ c

– 2 km

( )c

è ç ö

ø ÷

Rearrange, l 1,2 =-

è ç ö

2 km

ø ÷ k

è ç

ø ÷ ±

2 km

è ç

ø ÷

è ç ö

ø ÷ c

2 1–

ê ê

è ç

2 km

ø ÷

ú ú

Substitute:

and z=

Þ l 1,2 =-z wn

2 km

±z wn

è ç ö

ø ÷ c 1–

è ç

ø ÷

Þ l 1,2 =-z wn

± wn z2 1–

ë ê ù

è ç

ø ÷

û ú

Þ l 1,2 =-z wn

± wn z2

– 1

Solution Manual Engineering Vibration 5th Edition by Inman

Solution Manual Engineering Vibration 5th Edition by Inman – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Related

Be the first to review “Solution Manual Engineering Vibration 5th Edition by Inman” Cancel reply

solution manual Accounting Information Systems Richardson Chang Smith 3rd Edition – Updated 2024

Solution Manual Estimating in Building Construction 9th Edition By Peterson Dagostino

Solution Manual Advanced Transport Phenomena Analysis, Modeling, and Computations Ramachandran 1st edition – Updated 2024

Solution Manual Advanced Financial Accounting Christensen Cottrell Budd 13th edition – Updated 2024

Need Help

Explore

Categories

Solution Manual Engineering Vibration 5th Edition by Inman

Share this:

Solution Manual Engineering Vibration 5th Edition by Inman – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

Related

Be the first to review “Solution Manual Engineering Vibration 5th Edition by Inman” Cancel reply

Related products

solution manual Accounting Information Systems Richardson Chang Smith 3rd Edition – Updated 2024

Solution Manual Estimating in Building Construction 9th Edition By Peterson Dagostino

Solution Manual Advanced Transport Phenomena Analysis, Modeling, and Computations Ramachandran 1st edition – Updated 2024

Solution Manual Advanced Financial Accounting Christensen Cottrell Budd 13th edition – Updated 2024