14
OSCILLATIONS
Conceptual Questions
14.1. The period of a block oscillating on a spring is T  2 m/k  We are told that T1  20 s
(a) In this case the mass is doubled: m2  2m1
T2 2 m2 /k
m2
2m1



 2
T1 2 m1/k
m1
m1
So T2  2T1  2(20 s)  28 s
(b) In this case the spring constant is doubled: k2  2k1
T2 2 m/k2
k
k1
1

 1 

T1 2 m/k1
k2
2k1
2
So T2  T1/ 2  (20 s)/ 2  14 s
(c) The formula for the period does not contain the amplitude; that is, the period is independent of the amplitude.
Changing (in particular, doubling) the amplitude does not affect the period, so the new period is still 2.0 s.
It is equally important to understand what doesn’t appear in a formula. It is quite startling, really, the first time you
realize it, that the amplitude doesn’t affect the period. But this is crucial to the idea of simple harmonic motion. Of
course, if the spring is stretched too far, out of its linear region, then the amplitude would matter.
14.2. The period of a simple pendulum is T  2 L/g  We are told that T1  20 s
(a) In this case the mass is doubled: m2  2m1 However, the mass does not appear in the formula for the period of a
pendulum; that is, the period does not depend on the mass. Therefore the period is still 2.0 s.
(b) In this case the length is doubled: L2  2L1
T2 2 L2 /g
L2
2L1



 2
T1 2 L1/g
L1
L1
So T2  2T1  2(20 s)  28 s
(c) The formula for the period of a simple small-angle pendulum does not contain the amplitude; that is, the period is
independent of the amplitude. Changing (in particular, doubling) the amplitude, as long as it is still small, does not
affect the period, so the new period is still 2.0 s.
It is equally important to understand what doesn’t appear in a formula. It is quite startling, really, the first time you
realize it, that the amplitude (max ) doesn’t affect the period. But this is crucial to the idea of simple harmonic
motion. Of course, if the pendulum is swung too far, out of its linear region, then the amplitude would matter. The
amplitude does appear in the formula for a pendulum not restricted to small angles because the small-angle
approximation is not valid; but then the motion is not simple harmonic motion.
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14-1
14-2
Chapter 14
14.3. (a) The amplitude is the maximum displacement obtained. From the graph, A  10 cm
(b) From the graph, the period T  20 s The angular frequency is thus  
2
 31 rad/s
T
(c) The phase constant specifies at what point the cosine function starts. From the graph, the starting point looks like
1
1

A So at t  0, A cos  A means 0   (60) Since the oscillator is moving to the left at t  0, it is in the
2
2
3

upper half of the circular-motion diagram, and we choose  0   
3
14.4. (a) A position-vs-time graph plots x(t )  A cos(wt   0 ) The graph of x(t) starts at
1
A and is increasing. So
2
1


A  A cos0  0    We choose 0   since the particle is moving to the right, indicating that it is
2
3
3
in the bottom half of the circular-motion diagram.
at t  0,
(b) The phase at each point can be determined in the same manner as for part (a). For points 1 and 3, the amplitude is
1

again A At point 1, the particle is moving to the right so 1    At point 3, the particle is moving left, so
2
3

 3   . At point 2, the amplitude is A, so cos2  1   2  0
3
14.5. (a) A velocity-vs-time graph is a plot of v(t )   A sin (t   0 ) At t  0, the amplitude is
1
1
vmax  A
2
2

 1  7
 A sin  0 , so  0  sin 1    
or   Since v(t  0 s) is increasing, we need  sin (t  0 ) increasing at
2
6
6


7

t  0, so we choose  0 
6
1
5
1 
 At point 1, we need
(b) The phase is   wt   0  At points 1 and 3, v(t )   vmax , so   sin2 1    or
2
6
2 6
 sin  1, increasing, so  1 
5

 At point 3,  sin  3 is decreasing, so  3   At point 2, the amplitude is vmax , so
6
6

 sin  2  1, thus  2   .
2
1

14.6. Energy is transferred back and forth between all potential energy at the extremes  kA2  and all kinetic
2

1

energy at the equilibrium point(s)  mvmax 2   The equation does not say that the particle ever has amplitude A and
2

speed vmax at the same time. The equation relates expressions for the energy at two different times.
1
2
(b) The turning points occur where the potential energy is equal to the total energy. These points are at x  14 cm
and 26 cm.
14.7. (a) The equilibrium length is 20 cm since that is where the potential energy U  k ( x  xe )2  0
(c) The maximum kinetic energy of 7 J is obtained when the potential energy is minimal, at the equilibrium point.
(d) The total energy will be around 14 J. If a horizontal line is drawn at that energy, it intersects the potential energy
curve at the turning points x  12 cm and 28 cm.
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Oscillations
14-3
1
2
14.8. One expression for the total energy is E  kA2 , which occurs at the turning points when v(t )  0 If the total
energy is doubled,
1
1
2E  k ( A)2  k ( 2 A)2
2
2
Thus the new amplitude is A  2 A  2(20 cm)  28 cm
1
2
2
, which occurs at the equilibrium point when U  0 If the
14.9. One expression for the total energy is E  mvmax
total energy is doubled,
1
1
 )2  m( 2vmax )2
2E  m(vmax
2
2
  2vmax  2(20 cm/s)  28 cm/s
Thus the new maximum speed is vmax
14.10. The time constant  
m
decreases as b increases, meaning energy is removed from the oscillator more
b
quickly since E (t )  E0et/ .
(a) The medium is more resistive.
(b) The oscillations damp out more quickly.
(c) The time constant is decreased.
14.11. (a) T, the period, is the time for each cycle of the motion, the time required for the motion to repeat itself.
 , the damping time constant, is the time required for the amplitude of a damped oscillator to decrease to e1  37%
of its original value.
(b)  , the damping time constant, is the time required for the amplitude of a damped oscillator to decrease to
e1  37% of its original value, while t1/2 , the half-life, is the time required for the amplitude of a damped oscillator
to decrease to 50% of its original value.
14.12. Natural frequency is the frequency that an oscillator will oscillate at on its own. You may drive an oscillator
at a frequency other than its natural frequency. For example, if you are pushing a child in a swing, you build up the
amplitude of the oscillation by driving the oscillator at its natural frequency. You can achieve resonance by driving
an oscillator at its natural frequency.
Exercises and Problems
Section 14.1 Simple Harmonic Motion
14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence
T
1
1

 227 103 s  227 ms
f 440 Hz
14.2. Model: The air-track glider oscillating on a spring is in simple harmonic motion.
Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark.
33 s
 33s/oscillation  33 s
(a) T 
10 oscillations
1
1
 0303 Hz  030 Hz
(b) f  
T 33 s
(c)   2 f  2 0303 Hz  190rad/s
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14-4
Chapter 14
(d) The oscillation from one side to the other is equal to 60 cm  10 cm  50 cm  050 m Thus, the amplitude is
A  12 (050 m)  025 m
(e) The maximum speed is
 2 
vmax   A  
 A  (190 rad/s)(025 m)  048m/s
 T 
14.3. Model: The air-track glider attached to a spring is in simple harmonic motion.
Visualize: The position of the glider can be represented as x(t )  A cos t
Solve: The glider is pulled to the right and released from rest at t  0 s It then oscillates with a period T  20 s
and a maximum speed vmax  40cm/s  040m/s
2
2
v
040m/s

  rad/s  A  max 
 0127 m  127 cm  13 cm
T
20 s

 rad/s
(b) The glider’s position at t  025 s is
x025 s  (0127 m)cos[( rad/s)(025 s)]  0090 m  90 cm
(a) vmax   A and  
Section 14.2 Simple Harmonic Motion and Circular Motion
14.4. Model: The oscillation is the result of simple harmonic motion.
Solve: (a) The amplitude A  20 cm
(b) The period T  40 s, thus
1
1

 025 Hz
T 40 s
(c) The position of an object undergoing simple harmonic motion is x(t )  A cos(t   0) At t  0 s, x0 10 cm  Thus,
f 

 10 cm 
1  1 
10 cm (20 cm)cos 0  0  cos1 
  cos     rad  60.
3
 20 cm 
 2
Because the object is moving to the right at t  0 s, it is in the lower half of the circular motion diagram and thus
must have a phase constant between  and 2 radians. Therefore,  0  

3
rad  60
14.5. Model: The oscillation is the result of simple harmonic motion.
Solve: (a) The amplitude A  10 cm
(b) The time to complete one cycle is the period, hence T  20 s and
1
1
f  
 050 Hz
T 20 s
(c) The position of an object undergoing simple harmonic motion is x(t )  A cos(t   0) At t  0 s, x0  5 cm, thus
5 cm  (10 cm)cos[ (0 s)   0 ]
 cos 0 
5 cm
1
2
 1
    0  cos 1     
rad or  120
10 cm
2
2
3


Since the oscillation is originally moving to the left,  0  120
14.6. Visualize: The phase constant 23  has a plus sign, which implies that the object undergoing simple harmonic
motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left.
Solve: The position of the object is
x(t )  A cos(t   0 )  A cos(2 ft   0 )  (40 cm)cos[(4 rad/s)t  23  rad]
The amplitude is A  4 cm and the period is T  1/f  050 s A phase constant  0  2 /3 rad  120 (second
quadrant) means that x starts at  12 A and is moving to the left (getting more negative).
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Oscillations
14-5
Assess: We can see from the graph that the object starts out moving to the left.
14.7. Visualize: A phase constant of 

implies that the object that undergoes simple harmonic motion is in the
2
lower half of the circular motion diagram. That is, the object is moving to the right.
Solve: The position of the object is given by the equation
 

 
x(t )  A cos(t   0)  A cos(2 ft   0)  (80 cm)cos  rad/s  t  rad 
 2
 2

The amplitude is A  80 cm and the period is T  1/f  40 s With  0   /2 rad, x starts at 0 cm and is moving
to the right (getting more positive).
Assess: As we see from the graph, the object starts out moving to the right.
14.8. Solve: The position of the object is given by the equation
x(t )  Acos(t   0)  Acos(2 ft   0)
We can find the phase constant  0 from the initial condition:
0 cm  (40 cm)cos 0  cos 0  0   0  cos1(0)   12  rad
Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence,
 0   12  rad The final result, with f  40 Hz, is
x(t )  (40 cm)cos[(80 rad/s)t  12  rad]
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14-6
Chapter 14
14.9. Solve: The position of the object is given by the equation
x(t )  A cos(t   0)
The amplitude is A  80 cm The angular frequency   2 f  2 050 Hz   rad/s Since at t  0 it has its most
negative position, it must be about to move to the right, so  0    Thus
x(t )  80 cm cos[( rad/s)t   rad]
14.10. Model: The air-track glider is in simple harmonic motion.
Solve: (a) We can find the phase constant from the initial conditions for position and velocity:
x0  Acos 0
v0 x   Asin  0
Dividing the second by the first, we see that
sin  0
v
 tan  0   0 x
cos 0
 x0
The glider starts to the left ( x0  500 cm) and is moving to the right (v0 x  363 cm/s) With a period of 15 s  32 s,
the angular frequency is   2 /T  43  rad/s Thus

 1
363 cm/s
2
  3  rad (60) or  3  rad (  120)
 (4 /3 rad/s)(500 cm) 
 0  tan 1  
The tangent function repeats every 180°, so there are always two possible values when evaluating the arctan function.
We can distinguish between them because an object with a negative position but moving to the right is in the third
quadrant of the corresponding circular motion. Thus  0   23  rad, or 120
(b) At time t, the phase is    t   0  ( 43  rad/s)t  23  rad This gives    23  rad, 0 rad, 23  rad, and 43  rad at,
respectively, t  0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.
Section 14.3 Energy in Simple Harmonic Motion
Section 14.4 The Dynamics of Simple Harmonic Motion
14.11. Model: The block attached to the spring is in simple harmonic motion.
Solve: The period of an object attached to a spring is
T  2
m
 T0  20 s
k
where m is the mass and k is the spring constant.
(a) For mass  2m,
T  2
(b) For mass
2m
 ( 2)T0  28 s
k
1 m,
2
T  2
1m
2
k
 T0 / 2  141 s
(c) The period is independent of amplitude. Thus T  T0  20 s
(d) For a spring constant  2k ,
T  2
m
 T0 / 2  141 s
2k
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Oscillations
14-7
14.12. Model: The air-track glider attached to a spring is in simple harmonic motion.
Solve: Experimentally, the period is T  (120 s)/(10 oscillations)  120 s Using the formula for the period,
T  2
2
2
m
 2 
 2 
k 
 m
 (0200 kg)  548N/m
k
T


 120 s 
14.13. Model: The mass attached to the spring oscillates in simple harmonic motion.
Solve: (a) The period T  1/f  1/20 Hz  050 s
(b) The angular frequency   2 f  2  20 Hz  4 rad/s
(c) Using energy conservation
1 kA2
2
 12 kx02  12 mv02x
Using x0  50 cm, v0 x  30 cm/s and k  m 2  (0200 kg)(4 rad/s)2 , we get A  554 cm
(d) To calculate the phase constant  0,
 A cos  0  x0  50 cm
 50 cm 
  0  cos 1 
  045 rad
 554 cm 
(e) The maximum speed is vmax   A  (4 rad/s)(554 cm)  70cm/s
(f) The maximum acceleration is
amax   2 A   A  (4 rad/s)(70 cm/s)  88m/s2
2
 12 (0200 kg)(070 m/s)2  0049 J
(g) The total energy is E  12 mvmax
(h) The position at t  040 s is
x04 s  (554 cm)cos[(4 rad/s)(040 s)  045 rad]  38 cm
14.14. Model: The oscillating mass is in simple harmonic motion.
Solve: (a) The amplitude A  20 cm
(b) The period is calculated as follows:
2
2
 10rad/s  T 
 063 s
T
10 rad/s
(c) The spring constant is calculated as follows:
k

 k  m 2  (0050 kg)(10 rad/s)2  50N/m
m

(d) The phase constant  0   14  rad
(e) The initial conditions are obtained from the equations
x(t )  (20 cm)cos(10t  14  ) and vx (t )  (200 cm/s)sin (10t  14  )
At t  0 s, these equations become
x0  (20 cm)cos( 14  )  141 cm and v0 x  (20 cm/s)sin( 14  )  141cm/s

In other words, the mass is at 141 cm and moving to the right with a velocity of 14.1 cm/s.
(f) The maximum speed is vmax  A  (20 cm)(10 rad/s)  20cm/s
(g) The total energy E  12 kA2  12 (50 N/m)(0020 m)2  100 103 J
(h) At t  041 s, the velocity is
v0 x  (20 cm/s)sin[(10 rad/s)(040 s)  14  ]  146cm/s
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14-8
Chapter 14
14.15. Model: The block attached to the spring is in simple harmonic motion.
Visualize:
Solve: (a) The conservation of mechanical energy equation Kf  Usf  Ki  Usi is
1 mv 2
1
2
 12 k (x)2  12 mv02  0 J  0 J  12 kA2  12 mv02  0 J
m
10 kg
v0 
(040 m/s)  010 m  10 cm
k
16 N/m
(b) We have to find the velocity at a point where x  A/2 The conservation of mechanical energy equation
K2  Us2  Ki  Usi is
 A
2
1 2 1  A
1
1
1
11
11
 1
 3 1

mv2  k    mv02  0 J  mv22  mv02   kA2   mv02   mv02    mv02 
2
2 2
2
2
2
4 2
4 2
 2
 4 2

 v2 
3
3
v0 
(040 m/s)  0346m/s
4
4
The velocity is 35 cm/s.
Section 14.5 Vertical Oscillations
14.16. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
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Oscillations
14-9
Solve: (a) At equilibrium, Newton’s first law applied to the physics book is
( Fsp ) y  mg  0 N  k y  mg  0 N
 k  mg/y  (0500 kg)(98 m/s 2 )/(020 m)  245N/m  25 N/m
(b) To calculate the period:
k
245 N/m
2
2 rad

 70rad/s and T 

 090 s
m
0500 kg
 70 rad/s

(c) The maximum speed is
vmax  A  010 m70 rad/s  070m/s
Maximum speed occurs as the book passes through the equilibrium position.
14.17. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve: The period and angular frequency are
20 s
2
2
T
 06667 s and  

 9425rad/s
30 oscillations
T
06667 s
(a) The mass can be found as follows:

k
k
15 N/m
m 2 
 0169 kg  017 kg
m

(9425 rad/s)2
(b) The maximum speed vmax   A  (9425 rad/s)(0060 m)  057 m/s
14.18. Model: The vertical oscillations constitute simple harmonic motion.
Solve: To find the oscillation frequency using   2 f  k/m , we first need to find the spring constant k. In equilibrium,
the weight mg of the block and the spring force k L are equal and opposite. That is, mg  k L  k  mg/L The
frequency of oscillation f is thus given as
f 
1
2
k
1

m 2
mg/L
1

m
2
g
1

L 2
98 m/s 2
 35 Hz
0020 m
Section 14.6 The Pendulum
14.19. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve: The period of the pendulum is
T0  2
L0
 40 s
g
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14-10
Chapter 14
(a) The period is independent of the mass and depends only on the length. Thus T  T0  40 s
(b) For a new length L  2L0 ,
T  2
2 L0
 2T0  57 s
g
(c) For a new length L  L0/2,
L0 /2
1

T0  28 s
g
2
(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T  40 s
T  2
14.20. Model: Assume the small-angle approximation so there is simple harmonic motion.
Solve: The period is T  12 s/10 oscillations  120 s and is given by the formula
T  2
2
2
L
 T 
 120 s 
2
L
 g 
 (98 m/s )  36 cm
g
2

2





14.21. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve: (a) On the earth the period is
Tearth  2
L
10 m
 2
 20 s
g
980 m/s 2
(b) On Venus the acceleration due to gravity is
g Venus 
GM Venus
2
RVenus

(667  1011N  m 2 /kg 2 )(488  1024 kg)
 TVenus  2
(606  10 m)
6
2
 886m/s 2
L
10 m
 2
 21 s
g Venus
886 m/s 2
14.22. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple
harmonic motion.
Solve: Using the formula g  GM/R2 , the periods of the pendulums on the moon and on the earth are
Tearth  2
L
L
R2
L
R2
 2 earth earth and Tmoon  2 moon moon
g
GM earth
GM moon
Because Tearth  Tmoon ,
2
M
 R

L
R2
L
R2
2 earth earth  2 moon moon  Lmoon   moon  earth  Learth
GM earth
GM moon
 M earth  Rmoon 
2
 736  1022 kg  637  106 m 

(20 m)  33 cm
 598  1024 kg 
 174  106 m 



14.23. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion.
Solve: The time periods of the pendulums on the earth and on Mars are
Tearth  2
L
L
and TMars  2
gearth
gMars
Dividing these two equations,
Tearth

TMars
2
2
T

gMars
 150 s 
2
 g Mars  gearth  earth   (98 m/s 2 ) 
  367m/s
gearth
T
2

45
s


 Mars 
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Oscillations
14-11
14.24. Model: This is not a simple pendulum, but is a physical pendulum. Model the bar as thin enough to use
I  13 ML2 where L is the full length of the rod.
Visualize: For a small-angle physical pendulum T  2
I
Mgl
where l is the distance from the pivot to the center of
mass. Note that l = L/2. We seek L.
Solve: Substitute in for I and then solve for L.
T  2
1 ML2
I
 2 3
 2
Mgl
MgL/2
2
2 L2
3
gL
 2
2L
3g
2
 T  3g 3 g 2 3 (9.8 m/s )
L

T 
(1.2 s)2  0.54 m  54 cm

8
2
 2  2 8  2
Assess: It seems reasonable that a uniform bar 54 cm long would have a period of 1.2 s.
Section 14.7 Damped Oscillations
Section 14.8 Driven Oscillations and Resonance
14.25. Model: The spider is in simple harmonic motion.
Solve: Your tapping is a driving frequency. Largest amplitude at fext  10 Hz means that this is the resonance
frequency, so f0  fext  10 Hz That is, the spider’s natural frequency of oscillation f 0 is 1.0 Hz and  0  2 f0 
2 rad/s We have
 0
k
 k  m02  (00020 kg)(2 rad/s)2  0079N/m
m
14.26. Model: The motion is a damped oscillation.
Solve: The amplitude of the oscillation at time t is given by Equation 14.58: A(t )  A0et/2 , where   m/b is the
time constant. Using x  0368 A and t  100 s, we get
10 s
10.0 s
0.368 A  Ae10.0 s/2  ln(0.368) 
  
 5.00 s
2
2ln(0.368)
14.27. Model: The motion is a damped oscillation.
Solve: The position of a damped oscillator is x(t )  Ae(t/2 ) cos(t   0) The frequency is 1.0 Hz and the damping
time constant  is 4.0 s. Let us assume  0  0 rad and A  1 with arbitrary units. Thus,
x(t )  et/(80 s) cos[2 10 Hz t ]  x(t )  e0125 t cos(2 t )
where t is in s. Values of x(t) at selected values of t are displayed in the following table:
t(s)
0
0.25
0.50
0.75
1.00
1.25
1.50
1.75
x(t)
1
0
0.939
0
0.882
0
0.829
0
t(s)
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
x(t)
0.779
0.732
0.687
0.646
0.607
0.570
0.535
0.503
t(s)
6.00
6.50
7.00
7.50
8.00
8.50
9.00
9.50
10.00
x(t)
0.472
0.444
0.417
0.392
0.368
0.346
0.325
0.305
0.286
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14-12
Chapter 14
14.28. Model: The pendulum is a damped oscillator.
Solve: The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:
T  2
L
150 m
4(3600 s)
 2
 7773 s  Nosc 
 1850
2
g
7773 s
98 m/s
The amplitude of the pendulum as a function of time is A(t )  Aebt/2m  The exponent of this expression can be
calculated to be

bt
(0.010 kg/s)(4  3600 s)

 0.6545
2m
2(110 kg)
We have A(t )  150 me06545  078 m
14.29. Model: Assume the eye is a simple driven oscillator.
Visualize: Given the mass and the resonant frequency, we can determine the effective spring constant using the
relationship   2 f  k/m 
Solve: Solving the above expression for the spring constant, obtain
k  (2 f )2 m  [2  29 Hz ]2 (75 103 kg)  249 N/m  250 N/m
Assess: As spring constants go, this is a fairly large value, however the musculature holding the eyeball in the socket
is strong and hence will have a large effective spring constant.
14.30. Solve: The position and the velocity of a particle in simple harmonic motion are
x(t )  Acos(t   0 ) and vx (t )   A sin (t   0 )  vmax sin (t   0 )
From the graph, T  12 s and the angular frequency is
2 2 


 rad/s
T 12 s 6
(a) Because vmax  A  60cm/s, we have
A
60 cm/s
(b) At t  0 s,


60 cm/s
 /6rad/s
 115 cm
v0 x   A sin  0  30 cm/s   60 cm/s  sin  0  30 cm/s
  0  sin 1(05 rad)  16  rad (30) or 65  rad (150)
Because the velocity at t  0 s is negative and the particle is slowing down, the particle is in the second quadrant of
the circular motion diagram. Thus  0  65  rad
(c) At t  0 s, x0  (115 cm)cos( 65  rad)  100 cm
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Oscillations
14-13
14.31. Solve: The position and the velocity of a particle in simple harmonic motion are
x(t )  Acos(t   0) and vx (t )   A sin(t   0)  vmax sin(t   0)
(a) At t  0 s, the equation for x yields
(50 cm)  (100 cm)cos( 0)   0  cos1(05)   23  rad
Because the particle is moving to the left at t  0 s, it is in the upper half of the circular motion diagram, and the
phase constant is between 0 and  radians. Thus,  0  23  rad
(b) The period is 4.0 s. At t  0 s,
 2
v0 x   A sin  0  (100 cm) 
 T
  2
 sin 
  3

  136 cm/s

(c) The maximum speed is
 2 
vmax   A  
 (100 cm)  157cm/s
 40 s 
Assess: The negative velocity at t  0 s is consistent with the position-vs-time graph and the positive sign of the
phase constant.
14.32. Model: The vertical mass/spring systems are in simple harmonic motion.
Solve: Spring/mass A undergoes three oscillations in 12 s, giving it a period TA  40 s Spring/mass B undergoes 2
oscillations in 12 s, giving it a period TB  60 s We have
TA  2
 m  k  40 s 2
mA
mB
T
and TB  2
 A   A  B  

kA
kB
TB
 mB  kA  60 s 3
If mA  mB , then
kB 4
k
9
  A   225
kA 9
kB 4
14.33. Solve: The object’s position as a function of time is x(t )  Acos(t   0) Letting x  0 m at t  0 s, gives
0  A cos 0   0   12 
Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant
between  and 0 radians. Thus,  0   12  and
x(t )  A cos(t  12  )  x(t )  Asin t  (010 m)sin ( 12  t )
where we have used A  010 m and

Let us now find t where x  0060 m:
2 2 rad 

 rad/s
T
40 s
2
2
 
 0060 m 
0060 m  (010 m)sin  t   t  sin 1 
  041 s

2 
 010 m 
Assess: The answer is reasonable because it is approximately 18 of the period.
14.34. Model: The block attached to the spring is in simple harmonic motion.
Visualize: The position and the velocity of the block are given by the equations
x(t )  Acos(t   0 ) and vx (t )   A sin(t   0 )
Solve: (a) To graph x(t) we need to determine  ,  0 , and A. These quantities will be found by using the initial
(t  0 s) conditions on x(t) and vx (t ) The period is
T  2
m
10 kg
 2
 1405 s  14 s
k
20 N/m
 
2 2 rad

 4472rad/s
T 1405 s
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14-14
Chapter 14
At t  0 s, x0  A cos 0 and v0 x   A sin  0  Dividing these equations,
tan  0  
v0 x
 x0

(10m/s)
 11181   0  0841 rad
(4472 rad/s)(020 m)
From the initial conditions,
2
2
v 
 10m/s 
A  x02   0 x   (020 m)2  
  030 m
  
 4472 rad/s 
14.35. Model: The astronaut attached to the spring is in simple harmonic motion.
Solve: (a) From the graph, T  30 s, so we have
T  2
2
2
m
 T 
 30 s 
m
 k 
 (240 N/m)  55 kg
k
 2 
 2 
(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A  12 (080 m)  040 m,  0  0 rad, and

2
2

 21rad/s

T 30 s
The equation for the position of the astronaut is
x(t )  A cos t  10 m  (04 m) cos[(21 rad/s)t ]  10 m
 12 m  (04 m)cos[(21 rad/s)t ]  10 m  cos[(21 rad/s)t ]  05  t  050 s
The equation for the velocity of the astronaut is
vx (t )   A sin (t )
 v05 s  (04 m)(21 rad/s)sin[(21 rad/s)(050 s)]  073m/s
Thus her speed is 0.73 m/s.
14.36. Model: The particle is in simple harmonic motion.
Solve: The equation for the velocity of the particle is
vx (t )  (25 cm)(10 rad/s)sin(10 t )
Substituting into K  2U gives
1 2
1
 1
mvx (t )  2  kx 2 (t )   m[ (250 cm/s)sin (10 t )]2  k[(25 cm)cos(10 t )]2
2
2
 2

sin 2 (10 t )
2
 k  (25 cm)
 1  2
 2 
 2 2 
s
cos (10 t )
 m  (250 cm/s)2
 100 
2
1
 1  2
1
 tan 2 (10 t )  2(10 rad/s)2 
2.0  0.096 s
 s  2.0  t  tan
100
10


14.37. Model: The spring undergoes simple harmonic motion.
Solve: (a) Total energy is E  12 kA2  When the displacement is x  12 A, the potential energy is
U  12 kx2  12 k
 12 A
2
 14

1 kA2
2

1EK
4
 E  U  43 E
One quarter of the energy is potential and three-quarters is kinetic.
(b) To have U  12 E requires
U  12 kx 2  12 E  12

1 kA2
2
 x 
A
2
14.38. Solve: Average speed is vavg  x/t During half a period (t  12 T ), the particle moves from x   A to
x   A(x  2 A) Thus
vavg 
x 2 A 4 A
4A
2
2




 ( A)  vmax  vmax  vavg
t T/2 T
2 / 

2
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Oscillations
14-15
14.39. Model: The ball attached to a spring is in simple harmonic motion.
Solve: (a) Let t  0 s be the instant when x0  50 cm and v0  20 cm/s The oscillation frequency is

k
25 N/m

 50 rad/s
m
0100 kg
Solving Equation 14.26 for the amplitude gives
2
2
 20 cm/s 
v 
A  x02   0   (50 cm) 2  
  64 cm
 
 50 rad/s 
(b) The maximum acceleration is amax   2 A  160cm/s2 
(c) For an oscillator, the acceleration is most positive (a  amax ) when the displacement is most negative
( x   xmax   A) So the acceleration is maximum when x  64 cm
(d) We can use the conservation of energy between x0  50 cm and x1  30 cm:
1 mv 2
0
2
 12 kx02  12 mv12  12 kx12  v1  v02 
k 2
( x0  x12 )  0283m/s
m
The speed is 28 cm/s. Because k is known in SI units of N/m, the energy calculation must be done using SI units of
m, m/s, and kg.
14.40. Model: The block on a spring is in simple harmonic motion.
Solve: (a) The position of the block is given by x(t )  A cos(t   0) Because x(t )  A at t  0 s, we have
 0  0 rad, and the position equation becomes x(t )  Acost At t  0685 s, 300 cm  Acos(0685) and at
t  0886 s, 300 cm  Acos(0886) These two equations give
cos(0685 )   cos(0886 )  cos(  0886 )
 0685    0886    200rad/s
(b) Substituting into the position equation,
300 cm  A cos((200 rad/s)(0685 s))  Acos(1370)  020 A  A 
300 cm
 150 cm
020
14.41. Model: The oscillator is in simple harmonic motion. Energy is conserved.
Solve: The energy conservation equation E1  E2 is
1 mv 2
1
2
 12 kx12  12 mv22  12 kx22
1
1
1
1
(030 kg)(0954 m/s) 2  k (0030 m) 2  (030 kg)(0714 m/s)2  k (0060 m)2
2
2
2
2
 k  4448N/m
The total energy of the oscillator is
1
1
1
1
Etotal  mv12  kx12  (030 kg)(0954 m/s)2  (4448 N/m)(0030 m)2  01565 J
2
2
2
2
2
,
Because Etotal  12 mvmax
1
2
01565 J  (0300 kg)vmax
 v max  102m/s
2
Assess: A maximum speed of 1.02 m/s is reasonable.
14.42. Model: The transducer undergoes simple harmonic motion.
Solve: Newton’s second law for the transducer is
Frestoring  ma max  40,000 N  (010 103 kg)amax  amax  40 108m/s2
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14-16
Chapter 14
Because amax   2 A,
A
amax
2

40  108m/s 2
[2 10 106 Hz ]2
 101 105 m  101  m
(b) The maximum velocity is
vmax   A  2 10 106 Hz 101105 m  64m/s
14.43. Model: The block attached to the spring is in simple harmonic motion.
Solve: (a) The frequency is
f 
1
2
k
1

m 2
2000 N/m
 3183 Hz
50 kg
The frequency is 3.2 Hz.
(b) From energy conservation,
2
2
v 
 10 m/s 
A  x02   0   (0050 m)2  
  00707 m
 
 2  3183 Hz 
The amplitude is 7.1 cm.
(c) The total mechanical energy is
E  12 kA2  12 (2000 N/m)(00707 m)2  50 J
14.44. Model: Model this situation as a simple harmonic oscillator without damping.
Visualize: The period is independent of the amplitude. If the oscillator is undamped the amplitude will be constant;
the amplitude is just the arbitrary distance from the equilibrium from which the mass was released. The column of
amplitude data is not needed.
The period is related to the spring constant however: T  2
m
.
k
Solve: The equation T 2  4 2 mk leads us to believe that a graph of T 2 vs. m would give a straight line graph whose
slope is 4 2 /k . Note that the period is 1/10 of the times given in the table.
From the spreadsheet we see that the linear fit is very good and that the slope is 6.0997s2 /kg.
slope  4 2 /k  k 
4 2
6.0997s2 /kg
 6.5 N/m
Assess: Check the units carefully in the last equation. 6.5 N/m is a reasonable spring constant.
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Oscillations
14-17
14.45. Model: The block undergoes simple harmonic motion.
Visualize:
Solve: (a) The frequency of oscillation is
f 
1
2
k
1 10 N/m

 1125 Hz
m 2 020 kg
The frequency is 1.1 Hz.
(b) Using conservation of energy,
1 mv 2
1
2
 12 kx12  12 mv02  12 kx02 , we find
m 2 2
020 kg
(v0  v1 )  (020 m) 2 
((100 m/s) 2   050 m/s 2 )
k
10 N/m
 02345 m or 23 cm
(c) At time t, the displacement is x  A cos(t   0) The angular frequency is   2 f  7071 rad/s The amplitude is
x1  x02 
2
2
v 
 100 m/s 
A  x02   0   (020 m) 2  
  0245 m
 
 7071 rad/s 
The phase constant is
 x0 
1  0200 m 
  cos 
  2526 rad or  145
 A
 0245 m 
A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the
corresponding circular motion is in the third quadrant, so  0  2526 rad Thus at t  10 s,
 0  cos1 
x  (0245 m)cos((7071 rad/s)(10 s)  2526 rad)  00409 m  409 cm
The block is 4.1 cm below the equilibrium point.
14.46. Model: The mass is in simple harmonic motion.
Visualize:
The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational
potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches the spring, then the
elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back
to exactly its starting height. The total displacement of the oscillation—high point to low point—is 20 cm. Because the
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14-18
Chapter 14
oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm
below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm,
that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point.
Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp  FG  mg At the equilibrium
point, the spring is stretched by y  10 cm  010 m Hooke’s law is Fsp  k y, so the equilibrium condition is
k
g 98 m/s 2


 98 s2 2
m y 010 m
The ratio k/m is all we need to find the oscillation frequency:
[ Fsp  k y ]  [ FG  mg ] 
f 
1
2
k
1

98 s2  158 Hz  16 Hz
m 2
14.47. Model: The spring is ideal, so the apples undergo simple harmonic motion.
Solve: The spring constant of the scale can be found by considering how far the pan goes down when the apples
are added.
mg
mg
20 N
L 
k 

 222 N/m
k
L 0090 m
The frequency of oscillation is
f 
1
2
k
1

m 2
222 N/m
(20 N/98 m/s 2 )
 166 Hz  17 Hz
Assess: An oscillation of fewer than twice per second is reasonable.
14.48. Model: The compact car is in simple harmonic motion.
Solve: (a) The mass on each spring is (1200 kg)/4  300 kg The spring constant can be calculated as follows:
2 
k
 k  m 2  m(2 f )2  (300 kg)[2 (20 Hz)]2  474 104N/m
m
The spring constant is 47 104 N/m
(b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is,
m  300 kg  70 kg  370 kg Thus,
f 

1

2 2
k
1

m 2
474 104 N/m
 18 Hz
370 kg
Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional
to the square root of the mass.
14.49. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We
will also use the model of static friction between the two blocks.
Visualize:
Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2  The two blocks
ride together as long as the static friction doesn’t exceed its maximum possible value. The model of static friction
gives the maximum force of static friction as
( fs )max  s n  s (m1g )  m1amax  amax  s g
2
2
amax  2 Amax  2   Amax   2   040 m 



 
 
  072
g
g
 T   g   15 s   98 m/s 2 
Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation.
 s 
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Oscillations
14-19
14.50. Model: The DNA and cantilever undergo simple harmonic motion.
Solve: The cantilever has the same spring constant with and without the DNA molecule. The frequency of
oscillation without the DNA is
k
1 
1M
3
With the DNA, the frequency of oscillation is
2 
k
m
1M
3
where m is the mass of the DNA.
Divide the two equations, and express 2  1  , where   2f  2 (50 Hz)
1 f1
f1



2 f 2 f1  f


k
1M
3

k
m
1M
3


1M m
3
1M
3
Thus
f12
m  13 M
Since f
f1, (50 Hz
 13 M   ( f1  f )2  13 M  m 
( f12  ( f1  f ) 2 )
( f1  f )2


f12
 13 M 
1
 ( f  f ) 2 
 1

 f 
12 MHz), ( f1  f )2  f12 1 

f1 

2

f
 f12 1  2
f1


  Thus



f
f
m  13 M 1  2
 1  23 M
f1
f1


The mass of the cantilever
M  (2300 kg/m3 )(4000 109 m)(100 109 m)  368 1016 kg
Thus the mass of the DNA molecule is
2
 50 Hz 
21
m  (368  1016 kg) 
  102  10 kg
3
 12  106 Hz 
Assess: The mass of the DNA molecule is about 62 105 atomic mass units, which is reasonable for such a large
molecule.
14.51. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic
motion.
Visualize:
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14-20
Chapter 14
Solve: (a) Using the formula for the period of a pendulum,
T  2
2
2
L
 T 
2  55 s 
 L  g
  (98 m/s ) 
  75 m
g
2



 2 
(b) The conservation of mechanical energy equation K0  Ug0  K1  U g1 for the swinging lamp is
1 mv 2
0
2
2
 mgy0  12 mv12  mgy1  0 J  mgh  12 mvmax
0 J
 vmax  2 gh  2 g ( L  L cos3)
 2(98 m/s 2 )(75 m)(1  cos3)  045m/s
14.52. Model: Assume that the angle with the vertical that the pendulum makes is small enough so that there is
simple harmonic motion.
Solve: The angle  made by the string with the vertical as a function of time is
 (t )  max cos(t   0)
The pendulum starts from maximum displacement, thus  0  0 Thus,  (t )  max cos t To find the time t when the
pendulum reaches 4.0° on the opposite side:
(40)  80 cos t  t  cos1(05)  2094 rad
Using the formula for the angular frequency,

g
98 m/s2
20944 rad 2094 rad

 3130rad/s  t 

 0669 s
L
10 m

3130 rad/s
The time t  067 s
Assess: Because T  2 /  20 s, a value of 0.67 s for the pendulum to cover a little less than half the oscillation is
reasonable.
14.53. Model: Assume the orangutan is a simple small-angle pendulum.
Visualize: One swing of the arm would be half a period of the oscillatory motion. The horizontal distance traveled in
that time would be 2(090m)sin(20)  0616m from analysis of a right triangle.
Solve:
T 2

2
2
speed 
L 2

g
2
090 m
98 m/s 2
 0952s
dist 0616 m

 065 m/s
time 0952 s
Assess: This isn’t very fast, but isn’t out of the reasonable range.
14.54. Model: The mass is a particle and the string is massless.
Solve: Equation 14.51 is

Mgl
I
The moment of inertia of the mass on a string is I  Ml 2 , where l is the length of the string. Thus

Mgl
Ml 2

g
l
This is Equation 14.48 with L  l
Assess: Equation 14.48 is really a specific case of the more general physical pendulum described by Equation 14.51.
14.55. Model: The rod is thin and uniform with moment of inertia described in Table 12.2. The clay ball is a
particle located at the end of the rod. The ball and rod together form a physical pendulum. The oscillations are small.
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Oscillations
14-21
Visualize:
Solve: The moment of inertia of the composite pendulum formed by the rod and clay ball is
1
I rod  ball  I rod  I ball  mrod L2  mball L2
3
1
 (0200 kg)(015 m)2   0020 kg  015 m 2  195  103 kg m2
3
The center of mass of the rod and ball is located at a distance from the pivot point of
 015 
(0200 kg) 
m    0020 kg  015 m 
 2

ycm 
 818  102 m
(0200 kg  0020 kg)
The frequency of oscillation of a physical pendulum is
f 
1
2
The period of oscillation T 
Mgl
1

I
2
(0220 kg)(98 m/s 2 )(818 102 m)
195  103 kg m2
 151 Hz
1
 0.66 s.
f
14.56. Model: Model the rod as a small-angle physical pendulum without damping.
Visualize: Figure 14.22 in the chapter shows that l is the distance from the pivot to the center of mass; in this case
l  L/4.
Solve: We first need to compute the moment of inertia of a thin rod about an axis 1/4 of the way from one end. Use
the parallel-axis theorem to do so:
2
I  I c.m.  Md 2 
1
7
L  1 1 
ML2  M       ML2  ML2
12
48
 4   12 16 
For the physical pendulum
f 
1
2
Mgl
1

I
2
Mg
 L4  =
7 ML2
48
1 12 g 1 3 g
=
2 7 L  7 L
Assess: The period T  1/f is a little shorter than swinging the rod from the end, which is what we expect.
14.57. Model: Model the sphere as a small-angle physical pendulum without damping.
Visualize: Figure 14.22 in the chapter shows that l is the distance from the pivot to the center of mass; in this case l  R.
Solve: We first need to compute the moment of inertia of a sphere about an axis tangent to the edge. Use the
parallel-axis theorem to do so:
2
7
I  I c.m.  Md 2  MR 2  MR2  MR2
5
5
For the physical pendulum
1 Mgl
1
MgR
1 5g
f 


2
I
2 7 MR 2 2 7 R
5
Assess: The frequency is less than for a simple pendulum of length R as we expect.
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14-22
Chapter 14
14.58. Model: Model this situation as a small-angle physical pendulum without damping. The moment of inertia
for a uniform meter stick pivoted on one end is
1 ML2 .
3
Visualize: The frequency and length of a physical pendulum are related by f  2
Mgl
,
I
where l  L/2.
Solve: Square both sides of the equation above and substitute in for I.
f2
1 Mgl
1 Mg L2
3 g 3g 1

 2  2
2 I
2 1
2
4
4 3 ML 8 L 8 L
This leads us to believe that a graph of f 2 vs. 1/L would produce a straight line whose slope is
3g
.
8 2
From the spreadsheet we see that the linear fit is excellent and that the slope is 0.3707m/s2.
slope 
3g
8 2
g
(8 2 )(0.3707 m/s2 )
 9.76 m/s 2
3
Assess: If the uncertainties in our measurements were small enough then our answer would show that g varies
slightly from place to place on the earth.
14.59. Model: Treat the lower leg as a physical pendulum.
Visualize: We can determine the moment of inertia by combining T  2 I/mgL and T  1/f 
Solve: Combining the above expressions and solving for the moment of inertia we obtain
I  mgL/(2 f )2  (50 kg)(980 m/s2 )(018 m)/[2 (16 Hz)]2  87 102 kg  m2
Assess: NASA determines the moment of inertia of the shuttle in a similar manner. It is suspended from a heavy
cable, allowed to oscillate about its vertical axis of symmetry with a very small amplitude, and from the period of
oscillation one may determine the moment of inertia. This arrangement is called a torsion pendulum.
14.60. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion.
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Oscillations
14-23
Visualize:
Let us denote the 250 g and 500 g masses as m1 and m2 , which have initial velocities vi1 and vi2  After m1 collides
with and sticks to m2 , the two masses move together with velocity vf 
Solve: The momentum conservation equation pf  pi for the completely inelastic collision is (m1  m2 )vf  m1vi1  m2vi2.
Substituting the given values,
(0750 kg)vf  (0250 kg)(120 m/s)  (0500 kg)(0 m/s)  vf  0400m/s
We now use the conservation of mechanical energy equation:
( K  U s )compressed   K  U s equilibrium  0 J  12 kA2  12 (m1  m2 )vf2  0 J
 A
m1  m2
0750 kg
vf 
(0400 m/s)  011 m
k
10 N/m
The period is
T  2
m1  m2
0750 kg
 2
 17 s
k
10 N/m
14.61. Model: The block attached to the spring is oscillating in simple harmonic motion.
Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the
maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz.
(b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of mechanical
energy equation as follows:
1 kA2
2
2
 12 mvmax
 12 mv02
k
A   A  (2 f ) A  (2 )(20 Hz)(002 m)  025m/s
m
The blow to the block provides an impulse that changes the velocity of the block:
J x  Fx t  p  mvf  mv0
 v0 
(20 N)(10 103 s)  (0200 kg)vf  (0200 kg)(025 m/s)  vf  0150m/s
Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to A Thus,
A
0150m/s


0150m/s
 0012 m  119 cm  12 cm
2 (20 Hz)
Assess: Because vf is positive, the block continues to move to the right even after the blow.
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14-24
Chapter 14
14.62. Model: Assume the small-angle approximation.
Visualize:
Solve: The tension in the two strings pulls downward at angle   Thus Newton’s second law is
 Fy  2T sin  ma y
From the geometry of the figure we can see that
sin  
y
L  y2
2
If the oscillation is small, then y L and we can approximate sin  y/L Since y/L is tan  , this approximation
is equivalent to the small-angle approximation sin  tan if  1 rad With this approximation, Newton’s second
law becomes
2T sin   
2T
d2y
y  ma y  m 2
L
dt

d2y
dt 2

2T
y
mL
This is the equation of motion for simple harmonic motion (see Equations 14.32 and 14.46). The constants 2T/mL are
equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is
f 
1
2
2T
mL
14.63. Solve: The potential energy curve of a simple harmonic oscillator is described by U  12 k (x)2 , where
x  x  x0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0  0.13 nm.
We can find the bond’s spring constant by reading the value of the potential energy U at a displacement x and
using the potential energy formula to calculate k.
x (nm)
x (nm)
U (J)
k (N/m)
0.11
0.02
08 1019 J
400
0.10
0.03
19 1019 J
422
0.09
0.04
34 1019 J
425
The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring
constant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be
f 
1
2
k
1
416 N/m

 79  1013 Hz
m 2 167 1027 kg
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Oscillations
14-25
14.64. Model: Assume that the size of the ice cube is much less than R and that  is a small angle.
Visualize:
Solve: The ice cube is like an object on an inclined plane. The net force on the ice cube in the tangential direction is
 FG  sin   ma  mR  mR
d 2
 mg sin   mR
d 2
dt 2
dt 2
where  is the angular acceleration. With the small-angle approximation sin   , this becomes
d 2
g
     2
R
dt
This is the equation of motion of an object in simple harmonic motion with a period of
2
T
2

 2
R
g
14.65. Visualize:
Solve: (a) Newton’s second law applied to the penny along the y-axis is
Fnet  n  mg  ma y
Fnet is upward at the bottom of the cycle (positive a y ), so n  mg The speed is maximum when passing through
equilibrium, but a y  0 so n  mg The critical point is the highest point. Fnet points down and a y is negative. If
a y becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface.
(b) When the penny loses contact (n  0), the equation for Newton’s law becomes amax  g For simple harmonic
motion,
g
98 m/s 2

 1565 rad
A
0040 m
 1565 rad/s
f 

 25 Hz
2
2
amax  A 2   
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14-26
Chapter 14
14.66. Model: The vertical oscillations constitute simple harmonic motion.
Visualize:
Solve: At the equilibrium position, the net force on mass m on Planet X is:
k g
Fnet  k L  mgX  0 N   X
m L
For simple harmonic motion k/m   2 , thus
2 
2
2
gX
gX 2
 2 
 2

2
 

 gX  
 L  
 (0312 m)  586m/s
L
L T
T
14

5
s/10




Assess: Because ordinary tasks seemed easier than on earth we expected an answer less than 9.8 m/s2 .
14.67. Model: The doll’s head is in simple harmonic motion and is damped.
Solve: (a) The oscillation frequency is
1 k
 k  m(2 f )2  (0015 kg)(2 )2 (40 Hz) 2  9475N/m
2 m
The spring constant is 9.5 N/m.
f 
(b) Using A(t )  A0ebt/2m , we get
(05 cm)  (20 cm)eb(40 s)/(20015 kg)  025  e(1333 s/kg)b
 13333 s/kg b  ln 025  b  00104 kg/s  0010 kg/s
14.68. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement at time t of a damped oscillator is
t
 x (t ) 
xmax (t )  Aet/2    ln  max 
2
 A 
Using xmax  098 A at t  050 s, we can find the time constant  to be
050 s
 12375 s
2ln(098)
25 oscillations will be completed at t  25T  125 s At that time, the amplitude will be
 
xmax, 125 s  (10 cm)e125 s/(2)(12375 s)  60 cm
14.69. Model: The vertical oscillations are damped and follow simple harmonic motion.
Solve: The position of the ball is given by x(t )  Ae(t/2 ) cos(t   0) The amplitude A(t )  Ae(t/2 ) is a function
of time. The angular frequency is

k
(150 N/m)
2

 5477rad/s  T 
 1147 s
m
0500 kg

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Oscillations
14-27
Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 1147 s  34.41 s,
we have
3441 s
30 cm  (60 cm)e(34414 s/2 )  050  e(3441 s/2 )  ln(050) 
   25 s
2
14.70. Model: The motion is a damped oscillation.
Solve: The position of the air-track glider is x(t )  Ae(t/2 ) cos(t   0 ), where   m/b and

k
b2

m 4m 2
Using A  020 m,  0  0 rad, and b  0015 kg/s,

40N/m (0015 kg/s)2

 16  9  104 rad/s  40rad/s
0250 kg 4(0250 kg)2
Thus the period is
T
2


2 rad
 157 s
40 rad/s
The amplitude at t  0 s is x0  A and the amplitude will be equal to e1 A at a time given by
1
m
A  Ae(t/2 )  t  2  2  333 s
e
b
The number of oscillations in a time of 33.3 s is (333 s)/(157 s)  21
14.71. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement, or amplitude, of a damped oscillator decreases as xmax (t )  Aet/2 , where  is
the time constant. We know xmax /A  060 at t  50 s, so we can find  as follows:

t
50 s
 x (t ) 
 ln  max     
 489 s
2
2ln(060)
 A 
Now we can find the time t30 at which xmax /A  030:
 x (t ) 
t30  2 ln  max   2(489 s)ln(030)  118 s
 A 
The undamped oscillator has a frequency f  2 Hz  2 oscillations per second. Damping changes the oscillation
frequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, which allows
the oscillations to continue for well over 100 s, is certainly light damping, so we will use f  20 Hz Then the
number of oscillations before the spring decays to 30% of its initial amplitude is
N  f  t30  (2 oscillations/s)  (118 s)  236 oscillations
14.72. Solve: The solution of the equation
d 2x
dt 2

b dx k
 x0
m dt m
is x(t )  Aebt/2m cos(t   0 ). The first and second derivatives of x(t) are
dx
Ab bt/2m

e
cos(t   0 )  Aebt/2m sin (t   0 )
dt
2m


d 2 x  Ab 2
 Ab
  2  A 2  cos(t   0 ) 
sin (t   0 )  e bt/2 m
2


m
dt
 4m


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14-28
Chapter 14
Substituting these expressions into the differential equation, the terms involving sin (t   0 ) cancel and we obtain
the simplified result
 b2
k
2
 2     cos(t   0 )  0
m
 4m
Because cos(t   0 ) is not equal to zero in general,
b2
4m 2
 2 
k
k
b2
 0 

m
m 4m 2
14.73. Model: The two springs obey Hooke’s law.
Visualize:
Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces—
one pushing, the other pulling—are directed to the left and have negative values:
( Fnet ) x  ( Fsp 1) x  ( Fsp 2 ) x  k1x  k2 x  (k1  k2 ) x  keff x
where keff  k1  k2 is the effective spring constant. This means the oscillatory motion of the block under the influence
of the two springs will be the same as if the block were attached to a single spring with spring constant keff  The
frequency of the blocks, therefore, is
f 
1
2
keff
1

m
2
k1  k2
k1
k

 22 
2
m
4 m 4 m
f12  f 22
14.74. Model: The two springs obey Hooke’s law. Assume massless springs.
Visualize: Each spring is shown separately. Note that x  x1  x2 
Solve: Only spring 2 touches the mass, so the net force on the mass is Fm  F2 on m  Newton’s third law tells us that
F2 on m  Fm on 2 and that F2 on 1  F1 on 2  From Fnet  ma, the net force on a massless spring is zero. Thus Fw on 1
 F2 on 1  k1x1 and Fm on 2  F1 on 2  k2x2 Combining these pieces of information,
Fm  k1x1  k2x2
The net displacement of the mass is x  x1  x2 , so
x  x1  x2 
Fm Fm  1 1 
k k

    Fm  1 2 Fm
k1 k2  k1 k2 
k1k2
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Oscillations
14-29
Turning this around, the net force on the mass is
Fm 
k1k2
kk
x  keff x where keff  1 2
k1  k2
k1  k2
keff , the proportionality constant between the force on the mass and the mass’s displacement, is the effective spring
constant. Thus the mass’s angular frequency of oscillation is

keff
1 k1k2

m
m k1  k2
Using 12  k1/m and 22  k2 /m for the angular frequencies of either spring acting alone on m, we have

(k1/m)(k2 /m)
1222

(k1/m)  (k2 /m)
12  22
Since the actual frequency f is simply a multiple of  , this same relationship holds for f :
f 
f12 f 22
f12  f 22
14.75. Model: The blocks undergo simple harmonic motion.
Visualize:
The length of the stretched spring due to a block of mass m is L1 In the case of the two-block system, the spring is
further stretched by an amount L2 
Solve: The equilibrium equations from Newton’s second law for the single-block and double-block systems are
(L1)k  mg and (L1  L2 )k  (2m) g
Using L2  50 cm, and subtracting these two equations, gives us
(L1  L2 )k  L1k  (2m) g  mg  (005 m)k  mg
k 98 m/s2

 196 s2
m 005 m
With both blocks attached, giving total mass 2m, the angular frequency of oscillation is


k
1k
1


196 s2  990 rad/s
2m
2m
2
Thus the oscillation frequency is f  /2  16 Hz
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14-30
Chapter 14
14.76. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion.
Visualize:
Solve: (a) The equation for conservation of energy after the collision is
1 2 1
k
2500 N/m
kA  (mb  mB )vf2  vf 
A
(010 m)  50m/s
2
2
mb  mB
1010 kg
The momentum conservation equation for the perfectly inelastic collision pafter  pbefore is
(mb  mB )vf  mbvb  mBvB
(1010 kg)(50 m/s)  (0010 kg)vb  100 kg 0 m/s   vb  50 102m/s
(b) No. The oscillation frequency k/(mb  mB ) depends on the masses but not on the speeds.
14.77. Model: The block undergoes simple harmonic motion after sticking to the spring. Energy is conserved
throughout the motion.
Visualize:
It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward.
Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit on
the spring at rest. The spring is compressed by L at this point. Balancing the forces requires k L  mg The
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Oscillations
14-31
angular frequency is w2  k/m  g/L, so we can find the oscillation frequency by finding L The block hits the
spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been
transformed into the spring’s elastic energy. Equate the energies at these points:
K1  U1g  U3s  U3g  12 mv12  mg L  12 k (L  A)2  mg ( A)
We’ve used y1  L as the block hits and y3   A at the bottom. The spring has been compressed by y  L  A
Speed v1 is the speed after falling distance h, which from free-fall kinematics is v12  2 gh Substitute this expression
for v12 and mg/L for k, giving
mg
(L  A)2  mg ( A)
2(L)
The mg term cancels, and the equation can be rearranged into the quadratic equation
mgh  mg L 
(L)2  2h(L)  A2  0
The positive solution is
L  h2  A2  h  (0030 m)2   0100 m2  0030 m  00744 m
Now that L is known, we can find
g
980 m/s 2


 1148 rad/s  f 
 18 Hz
L
00744 m
2

14.78. Model: Assume that the pendulum is restricted to small angles so sin  
Visualize: Because the string is massless the center of mass is at the center of the sphere, so l  L in the formula for
a physical pendulum: T  2
I .
Mgl
Solve:
(a) First we use the parallel axis theorem to find I for the sphere around the pivot point at the top (L away):
I  Icm  Md 2   MR2  ML2.
T  2
2 MR 2
5
I
 2
Mgl
 ML2
MgL
This reduces to the traditional simple pendulum formula T  2
L
g
 2
2 R2
5
 L2
gL
in the limit L
R.
(b)
Treal

Tsimple
2
2 R2
5
 L2
gL
2
L
g

2 R2
5
L
 L2

2 (0.010 m) 2
5
 (1.0 m)2
1.0 m
 1.00002
Assess: Making the simple pendulum approximation introduces an error of only 0.002% with these numbers.
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14-32
Chapter 14
14.79. Model: Assume a simple system with a massless spring: T  2
m
.
k
Visualize: The simplest way to find T is to consider T as a function of m and use calculus to take the differential
dT and consider that a good stand-in for T . Also note that T  2
m
k
 k  4 2
m
.
T2
Solve:
(a) Re-write T  2
m
k
as a function of m.
2 1/2
m
k
Now take the differential of T with respect to m using the exponent rule and chain rule.
T ( m) 
dT 
Now cancel the 2s and plug in k  4 2
2  1  1/2
dm
 m
k 2
m
.
T2
dT 

1

dm 
k m
4 2
m
T2
1
T
dm 
dm
2m
m
Or, in the notation given in the problem:
T 
(b) The new period, T  , is given by
T   T  T  T 
T m
2 m
T m
 1 m 
 1

 T 1 
  (2.000 s) 1  (0.001)   2.001s
2 m
 2 m 
 2

Assess: We did not expect a 0.1% change in mass to change the period by much.
14.80. Model: The rod is thin and uniform.
Visualize:
Solve: We must derive our own equation for this combination of a pendulum and spring. For small oscillations,
Fs remains horizontal. The net torque around the pivot point is
L
 
 net  I   Fs L cos  FG   sin 
2
With  
d 2
1
, FG  mg , Fs  k x  kL sin , and I  mL2 ,
3
dt
2
d 2
dt
2

3k
3g
sin  cos 
sin 
m
2L
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Oscillations
14-33
1
We can use sin  cos  sin 2  For small angles, sin   and sin 2  2  So
2
d 2
dt 2
This is the same as Equations 14.32 and 14.46 with
 3k 3g 
  

 m 2L 

3k 3g

m 2L
The frequency of oscillation is thus
f 
1
2
3(30 N/m) 3(98 m/s 2 )

 173 Hz
(0200 kg) 2(020 m)
1
 0.58 s.
f
Assess: Fewer than two oscillations per second is reasonable. The rod’s angle from the vertical must be small
enough that sin 2  2  This is more restrictive than other examples, which only require that sin   
The period T 
© Copyright 2013 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.