CHAPTER 14: Oscillations
Responses to Questions
2.
The acceleration of a simple harmonic oscillator is momentarily zero as the mass passes through
the equilibrium point. At this point, there is no force on the mass and therefore no acceleration.
5.
The maximum speed of a simple harmonic oscillator is given by v  A
k
m
. The maximum speed
can be doubled by doubling the amplitude, A.
8.
The tire swing is a good approximation of a simple pendulum. Pull the tire back a short distance
and release it, so that it oscillates as a pendulum in simple harmonic motion with a small
amplitude. Measure the period of the oscillations and calculate the length of the pendulum from
the expression T  2
l
g
. The length, l , is the distance from the center of the tire to the branch.
The height of the branch is l plus the height of the center of the tire above the ground.
11. The two masses reach the equilibrium point simultaneously. The angular frequency is independent
of amplitude and will be the same for both systems.
14. When you rise to a standing position, you raise your center of mass and effectively shorten the
length of the swing. The period of the swing will decrease.
Solutions to Problems
3.
The spring constant is found from the ratio of applied force to displacement.
k
Fext
x

mg
x

 68 kg   9.80 m
3
5.0  10 m
s2
  1.333  10
5
N m
The frequency of oscillation is found from the total mass and the spring constant.
f 
4.
1
k
2
m

1
1.333  105 N m
2
1568 kg
 1.467 Hz  1.5 Hz
(a) The motion starts at the maximum extension, and so is a cosine. The amplitude is the
displacement at the start of the motion.
 2
T
x  A cos t   A cos 
 2 t   8.8cm cos 9.520t
 

 
 0.66 


t    8.8cm  cos 
  8.8cm  cos  9.5 t 
(b)
Evaluate the position function at t = 1.8 s.


x  8.8cm  cos 9.520s1 1.8s   1.252cm  1.3cm
5.
The period is 2.0 seconds, and the mass is 35 kg. The spring constant can be calculated from Eq. 147b.
T  2
m
k
 T 2  4 2
m
k
 k  4 2
m
T
2
 4 2
35kg
 2.0s 2
 350 N m
10. The spring constant is the same regardless of what mass is attached to the spring.
f 
1
2
k
m
k

4
2
 mf 2  constant  m1 f12  m2 f12 
 m kg  0.83 Hz    m kg  0.68 kg  0.60 Hz 
2
13.
2
 m
 0.68 kg  0.60 Hz 2
 0.83 Hz 2   0.60 Hz 2
(a)
For A, the amplitude is AA  2.5m . For B, the amplitude is AB  3.5m .
(b)
For A, the frequency is 1 cycle every 4.0 seconds, so f A  0.25 Hz .
For B, the frequency is 1 cycle every 2.0 seconds, so f B  0.50 Hz .
(c)
For C, the period is TA  4.0s . For B, the period is TB  2.0s
(d)
Object A has a displacement of 0 when t  0 , so it is a sine function.
xA  AAsin  2 f At  
xA   2.5m  sin  12  t 
Object B has a maximum displacement when t  0 , so it is a cosine function.
xB  ABcos  2 f Bt  
14.
xB   3.5m  cos  t 
Eq. 14-4 is x  Acos t    .
(a)
If x  0   A , then  A  A cos     cos1  1     .
 0.74 kg
(b)
If x  0  0 , then 0  A cos     cos1  0     12  .
(c)
If x  0  A , then A  A cos     cos1 1    0 .
(d)
If x  0  12 A , then
(e)
If x  0   12 A , then  12 A  A cos     cos1   12      23  .
(f)
If x  0  A
A  A cos     cos1  12      13  .
1
2
2 , then A
2  A cos     cos 1
 
1
2
    14  .
The ambiguity in the answers is due to not knowing the direction of motion at t = 0.
19. When the object is at rest, the magnitude of the spring force is equal to the force of gravity. This
determines the spring constant. The period can then be found.
F
 kx0  mg  k 
mg
T  2
m
x0
vertical
k
 2
m
 2
mg
x0
x0
g
0.14m
 2
9.80 m s 2
 0.75s
24. Consider the first free-body diagram for the block while it is
at equilibrium, so that the net force is zero. Newton’s
second law for vertical forces, with up as positive, gives this.
F
y
 FA  FB  mg  0  FA  FB  mg
r
FA
r
FB
r
FA
r
FB
x
r
mg
r
mg
Now consider the second free-body diagram, in which the
block is displaced a distance x from the equilibrium point. Each upward force will have increased
by an amount kx , since x  0 . Again write Newton’s second law for vertical forces.
F
y
 Fnet  FA  FB  mg  FA  kx  FB  kx  mg  2kx   FA  FB  mg   2kx
This is the general form of a restoring force that produces SHM, with an effective spring constant of
2k . Thus the frequency of vibration is as follows.
f 
28.
(a)
1
2
keffective m 
1
2k
2
m
The total energy is the maximum potential energy.
U  12 E 
1
2
kx 2 
1
2

1
2


xA
2  0.707 A
Now we are given that x  13 A.
(b)
U
E

1
2
kx 2
1
2
2
kA

x2
2
A

1
9
Thus the energy is divided up into
31.
kA2
1
9
potential and
8
9
kinetic .
The spring constant is found from the ratio of applied force to displacement.
k
F
x

95.0 N
0.175 m
 542.9 N m
Assuming that there are no dissipative forces acting on the ball, the elastic potential energy in the
loaded position will become kinetic energy of the ball.
Ei  E f

1
2
2
2
kxmax
 12 mvmax
 vmax  xmax
k
m
  0.175m 
542.9 N m
0.160 kg
 10.2 m s
35. (a) The work done in compressing the spring is stored as potential energy. The compressed
location corresponds to the maximum potential energy and the amplitude of the ensuing
motion.
W  12 kA2  k 
2W
2
A

2  3.6 J 
 0.13m 2
 426 N m  430 N m
(b) The maximum acceleration occurs at the compressed location, where the spring is exerting the
maximum force. If the compression distance is positive, then the acceleration is negative.
F  kx  ma  m  
kx
a

 426 N m  0.13m 
15m s2
 3.7 kg
41. The period of a pendulum is given by T  2 L g . The length is assumed to be the same for the
pendulum both on Mars and on Earth.
T  2 L g 
TMars  TEarth
g Earth
g Mars
TMars
TEarth

2 L g Mars
2 L g Earth
 1.35s 
1
0.37

g Earth

g Mars
 2.2 s
45. If we consider the pendulum as starting from its maximum displacement, then the equation of
2 t
motion can be written as   0 cos t  0 cos
. Solve for the time for the position to decrease
T
to half the amplitude.
1/ 2  12 0  0 cos
2 t1/ 2
T

2 t1/ 2
T

 cos 1 12 
3
 t1/ 2  16 T
It takes 16 T for the position to change from 10 to 5 . It takes 14 T for the position to change
from 10 to 0. Thus it takes 14 T  16 T  121 T for the position to change from 5 to 0. Due to the
symmetric nature of the cosine function, it will also take
1
12
T for the position to change from 0 to
5 , and so from 5 to 5 takes 16 T . The second half of the cycle will be identical to the first,
and so the total time spent between 5 and 5 is 13 T . So the pendulum spends one-third of its
time between 5 and 5.
47. Use energy conservation to relate the potential energy at the
l
maximum height of the pendulum to the kinetic energy at the
lowest point of the swing. Take the lowest point to be the zero
location for gravitational potential energy. See the diagram.

l cos 
h  l  l cos 
Etop  Ebottom  K top  U top  K bottom  U bottom 
2
 vmax  2 gh 
0  mgh  12 mvmax
2 gl 1  cos  
52. The meter stick used as a pendulum is a physical pendulum. The period is given by Eq. 14-14,
T  2
I
mgh
. Use the parallel axis theorem to find the moment of inertia about the pin. Express
the distances from the center of mass.
I  I CM  mh 
2
dT
dh
 2 
1
12
ml  mh
2
 l2

  121  h 
 h

1
2
1/ 2
2
 T  2
I
mgh
 2
 1l2 
  12 h 2  1  0  h 


1
12
ml 2  mh 2
mgh
1
12

2 

g
1
12
l
1/ 2

 h
h

2
l  0.2887 m
x  12 l  h  0.500  0.2887  0.211m from the end
Use the distance for h to calculate the period.
T
59.
(a)
and so
2  1 l
 12
g
1/ 2

 h
h

2
 1 1.00 m 2


 0.2887 m 

12
9.80 m s 2  0.2887 m

1/ 2
2
 1.53s
The energy of the oscillator is all potential energy when the cosine (or sine) factor is 1,

bt
E  kA  kA e m . The oscillator is losing 6.0% of its energy per cycle. Use this to find the
1
2
2
1
2
2
0
actual frequency, and then compare to the natural frequency.
E  t  T   0.94 E  t  
b
2m

1
2T
1
2
2
0
kA e

b t T 
m
bT

 1 2  btm 
 0.94  2 kA0 e   e m  0.94 


0
ln  0.94 
4
ln  0.94   
1
b2

2
2
02  2  0
ln  0.94  
ln  0.94  
f   f 0 2
b2


4
m
2

1

 1 2 2 1  1
1  1 2
1
0
f0
4m 0
16 2
16 2
2
ln  0.94 
 f   f0 
100   1.2  10 3  %

 f0 
2

1
2
16
2
 1.2  10 5  % diff  
(b) The amplitude’s decrease in time is given by A  A0e
solve for n. The value of
b
2m

was found in part (a).
bt
2m
. Find the decrease at a time of nT, and
A  A0e

bt
1
 A0e  A0e
2m
2
n
ln  0.94 

bnT
2m
 1
b
2m
nT  
1
2T
ln  0.94  nT 
 32.32  32 periods
60. The amplitude of a damped oscillator decreases according to A  A0e  t  A0e

bt
2m
. The data can be
used to find the damping constant.
A  A0e

bt
 b
2m
2m
t
 A0   2  0.075kg  ln  5.0   0.039 kg s

 
 3.5s 
 A
 2.0 
ln 
67. Apply the resonance condition,   0 , to Eq. 14-23, along with the given condition of
A0  23.7
F0
m
. Note that for this condition to be true, the value of 23.7 must have units of s 2 .
F0
A0 
m
A0   0  
 2  02   b2 2 m2
2
F0
m b 202 m 2


F0
F0
F
F
F

 0 2  Q 0  23.7 0
2
b
b
m0
k
k
m 0 m 0
m
m0
Q
 Q  23.7
81. Assume the block has a cross-sectional area of A. In the equilibrium position, the net force on the
block is zero, and so Fbuoy  mg. When the block is pushed into the water (downward) an additional
distance x, there is an increase in the buoyancy force ( Fextra ) equal to the weight of the additional
water displaced. The weight of the extra water displaced is the density of water times the volume
displaced.
Fextra  madd. g   waterVadd. g   water gAx    water gA x
water
water
This is the net force on the displaced block. Note that if the block is pushed down, the additional
force is upwards. And if the block were to be displaced upwards by a distance x , the buoyancy
force would actually be less than the weight of the block by the amount Fextra , and so there would
be a net force downwards of magnitude Fextra . So in both upward and downward displacement,
there is a net force of magnitude   water gA x but opposite to the direction of displacement. As a
vector, we can write the following.
r
r
Fnet     water gA x
This is the equation of simple harmonic motion, with a “spring constant” of k   water gA