Download Solutions to Problems

Solutions to Problems
1.
The particle would travel four times the amplitude: from x  A to x  0 to x   A to x  0 to
x  A . So the total distance  4 A  4  0.18 m   0.72 m .
2.
3.
The spring constant is the ratio of applied force to displacement.
F
180 N  75 N
105 N
k 

 5.3  10 2 N m
x 0.85 m  0.65 m 0.20 m
The spring constant is found from the ratio of applied force to displacement.
2
F mg  68 kg  9.8 m s
k 

 1.333 105 N m
3
x
x
5 10 m
The frequency of oscillation is found from the total mass and the spring constant.

f 
4.
(a)
1
k
2
m


1
1.333 105 N m
2
1568 kg
 1.467 Hz  1.5 Hz
The spring constant is found from the ratio of applied force to displacement.
2
F mg  2.7 kg  9.80 m s
k 

 735 N m  7.4 102 N m
x
x
3.6 102 m


(b)
The amplitude is the distance pulled down from equilibrium, so A  2.5 102 m
The frequency of oscillation is found from the total mass and the spring constant.
f 
5.
1
k
2
m

1
735 N m
2
2.7 kg
 2.626 Hz  2.6 Hz
The spring constant is the same regardless of what mass is hung from the spring.
1
f 
k m 
k 2  f m  constant  f1 m1  f 2 m2 
2
f 2  f1 m1 m2   3.0 Hz  0.60 kg 0.38 kg  3.8 Hz
6.
time
position
The table of data is
0
-A
shown, along with the
T
/4
0
smoothed graph.
T/2
A
Every quarter of a
3T/4
0
period, the mass
T
A
moves from an
5T/4
0
extreme point to the
equilibrium. The graph resembles a cosine wave (actually, the opposite of a cosine wave).
7.
The relationship between frequency, mass, and spring constant is f 
(a) f 
1
k
2
m

1
k
2
m
.

 k  4 2 f 2 m  4 2  4.0 Hz  2.5 104 kg  0.1579 N m  0.16 N m
2
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(b) f 
8.
1
k
2
m

1
0.1579 N m
2
5.0  10 4 kg
The spring constant is the same regardless of what mass is attached to the spring.
f 
1
k
2
m
k

4
 m kg  0.88 Hz 
9.
 2.8 Hz
2
2
 mf 2  constant  m1 f12  m2 f12 
  m kg  0.68 kg  0.60 Hz 
2

 0.68 kg  0.60 Hz 2
m
 0.88 Hz 2   0.60 Hz 2
 0.59 kg
(a) At equilibrium, the velocity is its maximum.
k
A   A  2 fA  2  3 Hz  0.13 m   2.450 m s  2.5 m s
m
(b) From Equation (11-5), we find the velocity at any position.
vmax 
 0.10 m 
v   vmax 1  2    2.45 m s  1 
 1.565 m
2
A
 0.13 m 
2
2
(c) Etotal  12 mvmax
 12  0.60 kg  2.45m s   1.801J  1.8 J
2
x2
s  1.6 m s
(d) Since the object has a maximum displacement at t = 0, the position will be described by the
cosine function.
x   0.13 m  cos  2  3.0 Hz  t  
x   0.13 m  cos  6.0 t 
10. The relationship between the velocity and the position of a SHO is given by Equation (11-5). Set
that expression equal to half the maximum speed, and solve for the displacement.
v  vmax 1  x 2 A2  12 vmax   1  x 2 A2 
1
2
 1  x 2 A2 
1
4
 x 2 A2 
3
4

x   3 A 2  0.866 A
11.
Since F   kx  ma for an object attached to a spring, the acceleration is proportional to the
displacement (although in the opposite direction), as a   x k m . Thus the acceleration will have
half its maximum value where the displacement has half its maximum value, at  12 x0
12.
on
The spring constant can be found from the stretch distance corresponding to the weight suspended
the spring.
k
F
mg
T  14 2 m k 

 2.62 kg   9.80 m s2 
 81.5 N m
x
x
0.315 m
After being stretched further and released, the mass will oscillate. It takes one-quarter of a period for
the mass to move from the maximum displacement to the equilibrium position.
1
4

 2.62 kg
 0.282 s
2 81.5 N m
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13.
the
(a)
The total energy of an object in SHM is constant. When the position is at the amplitude,
speed is zero. Use that relationship to find the amplitude.
Etot  12 mv 2  12 kx 2  12 kA2 
A
m
k
(b)
v2  x2 
3.0 kg
280 N m
 0.55 m s    0.020 m 
2
2
 6.034  102 m  6.0  10 2 m
Again use conservation of energy. The energy is all kinetic energy when the object has
its
maximum velocity.
2
Etot  12 mv 2  12 kx 2  12 kA2  12 mvmax

k
vmax  A
14.

280 N m
3.0 kg
 0.5829 m s  0.58 m s
The spring constant is found from the ratio of applied force to displacement.
F
80.0 N
k 
 4.00  102 N m
x 0.200 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
15.
m

 6.034  10 2 m
(a)

1
2
2
2
kxmax
 12 mvmax
 vmax  xmax
k
m
4.00 102 N m
  0.200 m 
0.180 kg
 9.43m s
The work done to compress a spring is stored as potential energy.
2  3.0 J 
2W
W  12 kx 2  k  2 
 416.7 N m  4.2  102 N m
2
x
 0.12 m 
(b) The distance that the spring was compressed becomes the amplitude of its motion. The
k
maximum acceleration is given by amax  A . Solve this for the mass.
m
 4.167  102 N m 
k
k
amax  A  m 
A
  0.12 m   3.333 kg  3.3 kg
m
amax
15 m s 2


16.
The general form of the motion is x  A cos  t  0.45 cos 6.40t .
(a)
The amplitude is A  xmax  0.45 m .
(b)
The frequency is found by   2 f  6.40 s 1  f 
(c)
The total energy is given by
2
Etotal  12 mvmax
 12 m  A 
2
(d)
1
2
6.40 s 1
2
 0.60 kg   6.40 s1   0.45 m 
2
 1.019 Hz  1.02 Hz
 2.488 J  2.5 J
The potential energy is given by
Epotential  12 kx 2  12 m 2 x 2 
1
2
 0.60 kg   6.40 s1   0.30 m 
2
2
 1.111J  1.1 J
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The kinetic energy is given by
Ekinetic  Etotal  Epotential  2.488 J  1.111 J  1.377 J  1.4 J
17.
total
If the energy of the SHO is half potential and half kinetic, then the potential energy is half the
energy. The total energy is the potential energy when the displacement has the value of the
amplitude.
Epot  12 Etot 
1
2
kx 2 
1
2

1
2
kA2


x
1
A  0.707 A
2
18. If the frequencies and masses are the same, then the spring constants for the two vibrations are the
same. The total energy is given by the maximum potential energy.
2
A 
A1
 1 2   1   7.0 
 7.0  2.6
E2 2 kA2  A2 
A2
E1
19.
(a)
pumpkin,
1
2
kA12
The general equation for SHM is Equation (11-8c), y  A cos  2 t T  . For the
 2 t  .

 0.65 s 
y   0.18 m  cos 
(b)
(c)
(d)
The time to return back to the equilibrium position is one-quarter of a period.
t  14 T  14  0.65 s   0.16 s
The maximum speed is given by the angular frequency times the amplitude.
2
2
vmax   A 
A
 0.18 m   1.7 m s
T
0.65 s
The maximum acceleration is given by
4 2
 2 
A

 0.18 m   17 m s 2 .

2
T 
 0.65 s 
2
amax   2 A  
The maximum acceleration is first attained at the release point of the pumpkin.
20. Consider the first free-body diagram for the block while it is
at equilibrium, so that the net force is zero. Newton’s 2nd
law for vertical forces, choosing up as positive, gives this.
 Fy  FA  FB  mg  0  FA  FB  mg
FA
FB
FA
FB
x
Now consider the second free-body diagram, in which the
mg
block is displaced a distance x from the equilibrium point.
mg
Each upward force will have increased by an amount  kx ,
since x  0 . Again write Newton’s 2nd law for vertical forces.
 Fy  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 
1
2
keffective m 
1
2k
2
m
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21.
The equation of motion is x  0.38sin 6.50 t  A sin t .
(a) The amplitude is A  xmax  0.38 m .
(b)
(c)
(d)
The frequency is found by   2 f  6.50 s 1  f 
 1.03Hz
2
The period is the reciprocal of the frequency. T  1 f  1 1.03 Hz  0.967 s .
The total energy is given by
2
Etotal  12 mvmax
 12 m  A 
2
(e)
6.50 s 1
1
2
 0.300 kg   6.50 s1   0.38 m 
2
 0.9151 J  0.92 J .
The potential energy is given by
Epotential  12 kx 2  12 m 2 x 2 
1
2
 0.300 kg   6.50 s1   0.090m 
2
2
 0.0513J  5.1 102 J .
The kinetic energy is given by
Ekinetic  Etotal  Epotential  0.9151J  0.0513J  0.8638 J  0.86 J .
(f)
0.4
x (m)
0.2
0
0
0.5
1
1.5
2
-0.2
-0.4
tim e (sec)
22.
(a)
For A, the amplitude is AA  2.5 m . For B, the amplitude is AB  3.5 m .
(b) For A, the frequency is 1 cycle every 4.0 seconds, so fA  0.25 Hz . For B, the frequency is 1
cycle every 2.0 seconds, so fB  0.50 Hz .
(c)
(d)
For C, the period is TA  4.0 s . For B, the period is TB  2.0 s
Object A has a displacement of 0 when t  0 , so it is a sine function.
xA  AA sin  2 f A t  
 
t
2 
xA   2.5 m  sin 
Object B has a maximum displacement when t  0 , so it is a cosine function.
xB  ABcos  2 f Bt  
23.
(a)
xB   3.5 m  cos  t 
Find the period and frequency from the mass and the spring constant.
T  2 m k  2 0.755 kg 124 N m   0.490 s
(b)
f  1 T  1  0.490 s   2.04 Hz
The initial speed is the maximum speed, and that can be used to find the amplitude.
vmax  A k m  A  vmax m k   2.96 m s  0.755 kg 124 N m   0.231 m
(c)
The maximum acceleration can be found from the mass, spring constant, and amplitude
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amax  Ak m   0.231 m 124 N m   0.755 kg   37.9 m s 2
(d)
Because the mass started at the equilibrium position of x = 0, the position function will be
proportional to the sine function.
x   0.231 m  sin  2  2.04 Hz  t  
(e)
position.
x   0.231 m  sin  4.08 t 
The maximum energy is the kinetic energy that the object has when at the equilibrium
2
E  12 mvmax
 12  0.755 kg  2.96 m s   3.31 J
2
24. We assume that downward is the positive direction of motion. For this motion, we have
k  305 N m , A  0.280 m , m  0.260 kg and   k m  305 N m 0.260 kg  34.250 rad s .
(a)
Since the mass has a zero displacement and a positive velocity at t = 0, the equation is a
sine
function.
(b)
y  t    0.280 m  sin  34.3rad s  t 
The period of oscillation is given by T 
2


2
34.25 rad s
 0.18345s . The spring will
have
its maximum extension at times given by the following.
T
tmax   nT  4.59  10 2 s  n  0.183 s  , n  0,1, 2,
4
The spring will have its minimum extension at times given by the following.
3T
tmin 
 nT  1.38  101 s  n  0.183 s  , n  0,1, 2,
4
25.
If the block is displaced a distance x to the right in the diagram, then spring # 1 will exert a force
F1  k1 x , in the opposite direction to x. Likewise, spring # 2 will exert a force F2  k2 x , in the
same direction as F1 . Thus the net force on the block is F  F1  F2  k1 x  k2 x    k1  k2  x . The
effective spring constant is thus k  k1  k2 , and the period is given by T  2
26.
m
k
 2
m
k1  k 2
.
The energy of the oscillator will be conserved after the collision. Thus
E  12 kA2 
1
2
2
 m  M  vmax
 vmax  A k  m  M 
This speed is the speed that the block and bullet have immediately after the collision. Linear
momentum in one dimension will have been conserved during the collision, and so the initial speed
of the bullet can be found.
pbefore  pafter  mvo   m  M  vmax
vo 
mM
m
A
k
mM

6.25  10 1 kg
2.5  10 2 kg
 2.15 10 m 
1
7.70  103 N m
6.25  10 1 kg
 597 m s
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27.
The period of the jumper’s motion is T 
38.0 s
8 cycles
 4.75 s . The spring constant can then be
found
from the period and the jumper’s mass.
T  2
m
k
4 2 m
 k
T2
4 2  65.0 kg 

 4.75 s 2
 113.73 N m  114 N m
The stretch of the bungee cord needs to provide a force equal to the weight of the jumper when he is
at the equilibrium point.
2
mg  65.0 kg  9.80 m s
k x  mg  x 

 5.60 m
k
113.73 N m


Thus the unstretched bungee cord must be 25.0 m  5.60 m  19.4 m
28.
(a)
(b)
60 s
The period is given by T 
 1.7 s cycle .
36 cycles
36 cycles
 0.60 Hz .
The frequency is given by f 
60 s
29. The period of a pendulum is given by T  2 L g . Solve for the length using a period of 2.0
seconds.
T  2 L g  L 
T 2g
4 2

 2.0 s 
2
 9.8 m s   0.99 m
2
4 2
30. 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
31.
g Earth
g Mars
TMars
TEarth

2 L g Mars
2 L g Earth
  0.80 s 
1
0.37

g Earth
g Mars

 1.3 s
The period of a pendulum is given by T  2 L g .
(a)
T  2 L g  2
0.80 m
 1.8 s
9.8 m s 2
If the pendulum is in free fall, there is no tension in the string supporting the pendulum
(b)
bob, and
so no restoring force to cause oscillations. Thus there will be no period – the pendulum will not
oscillate and so no period can be defined.
32. (a) The frequency can be found from the length of the
pendulum, and the acceleration due to gravity.
L
0
L cos
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
h  L  L cos
publisher.
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f 
1
g

9.80 m s 2
1
 0.57151 Hz  0.572 Hz
2 L 2 0.760 m
(b) To find the speed at the lowest point, use the conservation of
energy relating the lowest point to the release point of the pendulum. Take the lowest point to
be the zero level of gravitational potential energy.
Etop  Ebottom  KEtop  PEtop  KEbottom  PEbottom
2
0  mg  L  L cos  o   12 mvbottom
0

vbottom  2 gL 1  cos  o   2 9.80 m s 2
  0.760m  1  cos12.0   0.571 m s
o
(c) The total energy can be found from the kinetic energy at the bottom of the motion.
2
Etotal  12 mvbottom

1
2
 0.365 kg  0.571m s 
2
 5.95  10 2 J
33. There are  24 h  60 min h  60s min   86, 400 s in a day. The clock should make one cycle in
exactly two seconds (a “tick” and a “tock”), and so the clock should make 43,200 cycles per day.
After one day, the clock in question is 30 seconds slow, which means that it has made 15 less cycles
than required for precise timekeeping. Thus the clock is only making 43,185 cycles in a day.
43,185
Accordingly, the period of the clock must be decreased by a factor
.
43, 200
Tnew 
43,185
43, 200
 43,185 
 2 Lold g 
 43, 200 
Told  2 Lnew g  
2
Lnew
2
 43,185 
 43,185 

 Lold  
  0.9930 m   0.9923 m
 43, 200 
 43, 200 
Thus the pendulum should be shortened by 0.7 mm.
34. Use energy conservation to relate the potential energy at the
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.
Etop  Ebottom  KEtop  PEtop  KEbottom  PEbottom 
2
0  mgh  12 mvmax
 vmax  2 gh 
2 gL 1  cos  o 
L
0
L cos
h  L  L cos
35.
The equation of motion for an object in SHM that has the maximum displacement at t  0 is
given
by x  A cos  2 f t  . For a pendulum, x  L and so xmax  A  L max , where  must be measured
in radians. Thus the equation for the pendulum’s angular displacement is
L  L max cos  2 f t      max cos  2 f t 
If both sides of the equation are multiplied by 180o  rad , then the angles can be measured in
degrees. Thus the angular displacement of the pendulum can be written as below. Please note that
the argument of the cosine function is still in radians.
o
 o   max
cos  2 ft   15o cos  5.0 t 
(a)  o  t  0.25 s   15o cos  5.0  0.25   11o
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(b)  o  t  1.6 s   15o cos  5.0 1.6    15o
(here the time is exactly 4 periods)
(c)  o  t  500 s   15o cos  5.0  500    15o
36.
(here the time is exactly 1250 periods)
The wave speed is given by v   f . The period is 3.0 seconds, and the wavelength is 6.5 m.
v   f   T   6.5 m   3.0 s   2.2 m s
37.
The distance between wave crests is the wavelength of the wave.
  v f  343m s 262 Hz  1.31 m
38. To find the wavelength, use   v f .
AM:
v 3.00  108 m s
v 3.00  108 m s
1  

545
m



 188 m
2
f1
550  103 Hz
f 2 1600  103 Hz
FM:
1 
v
f1

3.00  108 m s
88.0  106 Hz
 3.41 m
2 
v

f2
3.00 108 m s
108 106 Hz
AM: 190 m to 550 m
 2.78 m
FM: 2.78 m to 3.41 m
39. The elastic and bulk moduli are taken from Table 9-1 in chapter 9. The densities are taken from
Table 10-1 in chapter 10.
v B  
(a) For water:
(b)
For granite:
(c)
For steel:
2.0  109 N m 2
1.00  10 kg m
3
v E  
3
 1.4  103 m s
45  109 N m 2
2.7  10 kg m
3
v E  
3
 4.1 103 m s
200 109 N m 2
7.8  103 kg m3
 5.1 103 m s
40. The speed of a longitudinal wave in a solid is given by v  E  . Call the density of the less dense
material 1 , and the density of the more dense material  2 . The less dense material will have the
higher speed, since the speed is inversely proportional to the square root of the density.
v1
v2

E 1
E 2

2
 2  1.41
1
41. To find the time for a pulse to travel from one end of the cord to the other, the velocity of the pulse
on the cord must be known. For a cord under tension, we have v 
FT
m L
.
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v
x
t
FT

m L
x
 t 
28 m

FT
 0.35 s
150 N
 0.65 kg   28 m 
m L
42. (a) The speed of the pulse is given by
x 2  620 m 
v

 77.5 m s  78 m s
t
16 s
(b) The tension is related to the speed of the pulse by v 
FT
m L
. The mass per unit length of the
cable can be found from its volume and density.
2
2
d
3
3  1.5  10 m 
 

       7.8  10 kg m  
  1.378 kg m
2
V   d 2 L
L
2
2


m
v
m
FT
m L
2
m
 FT  v 2
m
L
  77.5 m s  1.378 kg m   8.3 103 N
2
43. The speed of the water wave is given by v  B  , where B is the bulk modulus of water, from
Table 9-1, and  is the density of sea water, from Table 10-1. The wave travels twice the depth of
the ocean during the elapsed time.
v
2L
t
 L
vt
2

t
B
2


3.0 s
2.0  109 N m 2
2
1.025 10 kg m
3
3
 2.1 103 m
44. (a) Both waves travel the same distance, so x  v1 t1  v2 t2 . We let the smaller speed be v1 , and
the larger speed be v2 . The slower wave will take longer to arrive, and so t1 is more than t2 .
t1  t2  2.0 min  t2  120 s  v1  t2  120 s   v2t2 
t2 
v1
v2  v1
120 s  
5.5 km s
8.5 km s  5.5km s
120 s   220 s
x  v2t2   8.5 km s  220 s   1.9  103 km
(b) This is not enough information to determine the epicenter. All that is known is the distance of
the epicenter from the seismic station. The direction is not known, so the epicenter lies on a
circle of radius 1.9  103 km from the seismic station. Readings from at least two other seismic
stations are needed to determine the epicenter’s position.
45. We assume that the earthquake wave is moving the ground vertically, since it is a transverse wave.
An object sitting on the ground will then be moving with SHM, due to the two forces on it – the
normal force upwards from the ground and the weight downwards due to gravity. If the object loses
contact with the ground, then the normal force will be zero, and the only force on the object will be
its weight. If the only force is the weight, then the object will have an acceleration of g downwards.
Thus the limiting condition for beginning to lose contact with the ground is when the maximum
acceleration caused by the wave is greater than g. Any larger downward acceleration and the ground
would “fall” quicker than the object. The maximum acceleration is related to the amplitude and the
frequency as follows.
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amax   2 A  g  A 
g
2

g
4 2 f 2
9.8 m s 2

4 2  0.50 Hz 
2
 0.99 m
46. (a) Assume that the earthquake waves spread out spherically from the source. Under those
conditions, Eq. (11-16b) applies, stating that intensity is inversely proportional to the square of
the distance from the source of the wave.
I 20 km I10 km  10 km  20 km   0.25
(b) The intensity is proportional to the square of the amplitude, and so the amplitude is inversely
proportional to the distance from the source of the wave.
2
2
A20 km A10 km  10 km 20 km  0.50
47. (a) Assuming spherically symmetric waves, the intensity will be inversely proportional to the
square of the distance from the source. Thus Ir 2 will be constant.
2
I near rnear
 I far rfar2 
I near  I far
rfar2
2
near
r

 2.0  106 W m 2
48 km 

2
1 km 
2
 4.608  109 W m 2  4.6  109 W m 2
(b) The power passing through an area is the intensity times the area.



P  IA  4.608 109 W m2 5.0 m 2  2.3 1010 W
48. From Equation (11-18), if the speed, medium density, and frequency of the two waves are the same,
then the intensity is proportional to the square of the amplitude.
I 2 I1  E2 E1  A22 A12  2  A2 A1  2  1.41
The more energetic wave has the larger amplitude.
49. From Equation (11-18), if the speed, medium density, and frequency of the two waves are the same,
then the intensity is proportional to the square of the amplitude.
I 2 I1  P2 P1  A22 A12  3  A2 A1  3  1.73
The more energetic wave has the larger amplitude.
50. The bug moves in SHM as the wave passes. The maximum KE of a particle in SHM is the total
energy, which is given by Etotal  12 kA2 . Compare the two KE maxima.
2
 A   2.25 cm 
 1 2  2  
  0.56
KE1 2 kA1  A1   3.0 cm 
1
2
KE2
kA22
2
51. (a)
(b)
1
1
0.5
0.5
0
-0.5
0
-0.5
-1
-1
-4
-2
0
2
4
-4
-2
0
2
4
(c) The energy is all kinetic energy at the moment when the string has no displacement. There is
no elastic potential energy at that moment. Each piece of the string has speed but no
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displacement.
52. The frequencies of the harmonics of a string that is fixed at both ends are given by f n  nf1 , and so
the first four harmonics are f1  440 Hz , f 2  880 Hz , f3  1320 Hz , f 4  1760 Hz .
53. The fundamental frequency of the full string is given by f unfingered 
v
 294 Hz . If the length is
2L
reduced to 2/3 of its current value, and the velocity of waves on the string is not changed, then the
new frequency will be
v
3 v  3
 3
f fingered  2

   f unfingered    294 Hz  441 Hz
2  3 L  2 2L  2 
 2
54. Four loops is the standing wave pattern for the 4th harmonic, with a frequency given by
f 4  4 f1  280 Hz . Thus f1  70 Hz , f 2  140 Hz , f3  210 Hz and f5  350 Hz are all other
resonant frequencies.
55. Adjacent nodes are separated by a half-wavelength, as examination of Figure 11-40 will show.
v
v
92 m s

 xnode  12  

 9.7  102 m
f
2 f 2  475 Hz 
56. Since f n  nf1 , two successive overtones differ by the fundamental frequency, as shown below.
f  fn1  fn   n  1 f1  nf1  f1  350 Hz  280 Hz  70 Hz
57. The speed of waves on the string is given by equation (11-13), v 
of a string with both ends fixed are given by equation (11-19b), f n 
FT
m L
. The resonant frequencies
nv
, where Lvib is the length
2 Lvib
of the portion that is actually vibrating. Combining these relationships allows the frequencies to be
calculated.
fn 
n
2 Lvib
FT
m L
f 2  2 f1  581.54 Hz
f1 
1
2  0.62 m 
 3.6 10
520 N
3
kg
  0.90 m 
 290.77 Hz
f3  3 f1  872.31Hz
So the three frequencies are 290 Hz , 580 Hz , 870 Hz , to 2 significant figures.
58. From Equation (11-19b), f n 
nv
2L
, we see that the frequency is proportional to the wave speed on
the stretched string. From equation (11-13), v 
FT
, we see that the wave speed is proportional
m L
to the square root of the tension. Thus the frequency is proportional to the square root of the tension.
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FT 2

FT 1
2
f 
 200 Hz 
 FT 2   2  FT 1  
 FT 1  0.952 FT 1
 205 Hz 
 f1 
f2
f1
2
Thus the tension should be decreased by 4.8% .
59. The string must vibrate in a standing wave pattern to have a certain number of loops. The frequency
of the standing waves will all be 60 Hz, the same as the vibrator. That frequency is also expressed
nv
by Equation (11-19b), f n 
. The speed of waves on the string is given by Equation (11-13),
2L
v
FT
. The tension in the string will be the same as the weight of the masses hung from the end
m L
of the string, FT  mg . Combining these relationships gives an expression for the masses hung
from the end of the string.
nv
(a) f n 
2L

FT
n

2L m L
n
mg
2L
m L
5

1 9.80 m s
2

2

2
m1
(c) m5 
2
2
m1
(b) m2 

4 1.50 m   60 Hz  3.9 104 kg m
2
m1 
 m
1.289 kg
4
1.289 kg
25
2

4 L2 f n2  m L 
n2 g
  1.289 kg  1.3 kg
 0.32 kg
 5.2  102 kg
60. The tension in the string is the weight of the hanging mass, FT  mg . The speed of waves on the
FT
mg
, and the frequency is given as f  60 Hz . The
m L
m L
wavelength of waves created on the string will thus be given by
string can be found by v 

v
f

1
mg
f
m L

1
60 Hz

 0.080 kg   9.80 m s 2 
 3.9 10
4
kg m

 0.7473 m .
The length of the string must be an integer multiple of half of the wavelength for there to be nodes at
both ends and thus form a standing wave. Thus L   2, , 3 2, , and so on. This gives
L  0.37 m , 0.75 m , 1.12 m , 1.49 m as the possible lengths, and so there are 4 standing wave
patterns that may be achieved.
61. From the description of the water’s behavior, there is an anti-node at each end of the tub, and a node
in the middle. Thus one wavelength is twice the tube length.
v   f   2 Ltub  f  2  0.65 m  0.85 Hz   1.1m s
62. The speed in the second medium can be found from the law of refraction, Equation (11-20).
sin  2 v2
sin  2
 sin 35o 

 v2  v1
  8.0 km s  
 6.3km s
o 
sin 1 v1
sin 1
 sin 47 
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63. The angle of refraction can be found from the law of refraction, Equation (11-20).
sin  2 v2
v
2.1m s

 sin  2  sin 1 2  sin 34o
 0.419   2  sin 1 0.419  25o
sin 1 v1
v1
2.8 m s
64. The angle of refraction can be found from the law of refraction, Equation (11-20). The relative
velocities can be found from the relationship given in the problem.
331  0.60  10 
sin  2 v2 331  0.60T2
325
 
 sin  2  sin 25o
 sin 25o
 0.4076
sin 1 v1 331  0.60T1
331  0.60 10 
337
 2  sin 1 0.4076  24o
65. The angle of refraction can be found from the law of refraction, Equation (11-20). The relative
velocities can be found from Equation (11-14a).
sin  2
sin 1

v2
v1

sin  2  sin 1
E 2
E 1
SG1
SG2

SG1  water
1
SG1


2
SG2  water
SG2
 sin 38o
3.6
2.8
 0.70   2  sin 1 0.70  44o
66. The error of 2o is allowed due to diffraction of the waves. If the waves are incident at the “edge” of
the dish, they can still diffract into the dish if the relationship    L is satisfied.

 rad 

   L   0.5 m   2o 
 1.745  102 m  2  10 2 m
o 
L
180 

If the wavelength is longer than that, there will not be much diffraction, but “shadowing” instead.

67. The unusual decrease of water corresponds to a trough in Figure 11-24. The crest or peak of the
wave is then one-half wavelength distant. The peak is 125 km away, traveling at 750 km/hr.
x
125 km  60 min 
x  vt  t 


  10 min
v 750 km hr  1 hr 
68. Apply the conservation of mechanical energy to the car, calling condition # 1 to be before the
collision and condition # 2 to be after the collision. Assume that all of the kinetic energy of the car is
converted to potential energy stored in the bumper. We know that x1  0 and v2  0 .
E1  E2 
x2 
m
k
v1 
1
2
mv12  12 kx12  12 mv22  12 kx22 
1500 kg
550  103 N m
 2.2 m s  
1
2
mv12  12 kx22 
0.11 m
69. Consider the conservation of energy for the person. Call the unstretched position of the fire net the
zero location for both elastic potential energy and gravitational potential energy. The amount of
stretch of the fire net is given by x, measured positively in the downward direction. The vertical
displacement for gravitational potential energy is given by the variable y, measured positively for the
upward direction. Calculate the spring constant by conserving energy between the window height
and the lowest location of the person. The person has no kinetic energy at either location.
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2
Etop  Ebottom  mgytop  mgybottom  12 kxbottom
k  2mg
y
top
 ybottom 
2
bottom
x

 2  65 kg  9.8 m s 2

18 m   1.1 m  
1.1 m 
2
 2.011 104 N m
(a) If the person were to lie on the fire net, they would stretch the net an amount such that the
upward force of the net would be equal to their weight.
2
mg  65 kg  9.8 m s
F  kx  mg  x 

 3.2  102 m
4
k
2.01110 N m
(b) To find the amount of stretch given a starting height of 35 m, again use conservation of energy.
Note that ybottom   x , and there is no kinetic energy at the top or bottom positions.


Etop  Ebottom  mgytop  mgybottom  12 kx 2  x 2  2
 65 kg   9.8 m
x 2
2
2.011  10 4
mg
k
x2
 x  2  65 kg   9.8 m s   35 m   0
N m
2.011  10 N m
s2
mg
k
ytop  0
2
4

x 2  0.06335 x  2.2173  0  x  1.5211 m ,  1.458 m
This is a quadratic equation. The solution is the positive root, since the net must be below the
unstretched position. The result is 1.5 m .
70. Consider energy conservation for the mass over the range of motion
from “letting go” (the highest point) to the lowest point. The mass falls
the same distance that the spring is stretched, and has no KE at either
endpoint. Call the lowest point the zero of gravitational potential
energy. The variable “x” represents the amount that the spring is
stretched from the equilibrium position.
Etop  Ebottom
1
2
f 

1

2 2
2g
H

k

m

1
2 9.8 m s 2
2
0.33 m
2g
H
y=H
x=H
2
2
2
2
mvtop
 mgytop  12 kxtop
 12 mvbottom
 mgybottom  12 kxbottom
0  mgH  0  0  0  12 kH 2 
x=0
 2   
y=0
2g
H
  1.2 Hz
71. (a) From conservation of energy, the initial kinetic energy of the car will all be changed into elastic
potential energy by compressing the spring.
E1  E2  12 mv12  12 kx12  12 mv22  12 kx22  12 mv12  12 kx22 
km
v12
2
2
x
 22 m s 
  950 kg 
2
 5.0 m 
2
 1.8392  10 4 N m  1.8  10 4 N m
(b) The car will be in contact with the spring for half a period, as it moves from the equilibrium
location to maximum displacement and back to equilibrium.
1
2
T  12 2
m
k

 950 kg 
1.8392  104 N m
 0.71 s
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72. The frequency at which the water is being shaken is about 1 Hz. The sloshing coffee is in a standing
wave mode, with anti-nodes at each edge of the cup. The cup diameter is thus a half-wavelength, or
  16 cm . The wave speed can be calculated from the frequency and the wavelength.
v   f  16 cm 1 Hz   16 cm s
73. Relative to the fixed needle position, the ripples are moving with a linear velocity given by
 rev  1min   2  0.108 m  
v   33


  0.373 m s
1 rev
 min  60 s  

This speed is the speed of the ripple waves moving past the needle. The frequency of the waves is
v
0.373 m s
f  
 220 Hz
 1.70  103 m
74. The equation of motion is x  0.650cos 7.40 t  A cos t .
(a) The amplitude is A  0.650 m
(b) The frequency is given by   2 f  7.40 rad s  f 
7.40 rad s
2 rad
 1.177 Hz  1.18 Hz
(c) The total energy is given by
Etotal  12 kA2  12 m 2 A2 
(d) The potential energy is found by
PE  12 kx2  12 m 2 x2 
The kinetic energy is found by
1
2
 2.00 kg  7.40 rad s   0.650 m
1
2
 2.00 kg  7.40 rad s   0.260 m 
2
2
2
 23.136 J  23.1 J .
2
 3.702 J  3.70 J .
KE  Etotal  PE  23.136 J  3.702 J  19.4 J .
75. The frequency of a simple pendulum is given by f 
1
g
. The pendulum is accelerating
2 L
vertically which is equivalent to increasing (or decreasing) the acceleration due to gravity by the
acceleration of the pendulum.
(a) f new 
(b) f new 
1
ga
2
L
1
ga
2
L


1
1.50 g
2
L
1
0.5 g
2
L
 1.50
 0.5
1
g
2
L
1
g
2
L
 1.50 f  1.22 f
 0.5 f  0.71 f
76. The force of the man’s weight causes the raft to sink, and that causes the water to put a larger upward
force on the raft. This extra buoyant force is a restoring force, because it is in the opposite direction
of the force put on the raft by the man. This is analogous to pulling down on a mass-spring system
that is in equilibrium, by applying an extra force. Then when the man steps off, the restoring force
pushes upward on the raft, and thus the raft – water system acts like a spring, with a spring constant
found as follows.
2
F  75 kg  9.8 m s
k 
 1.8375 104 N m
2
x
4.0 10 m
(a) The frequency of vibration is determined by the spring constant and the mass of the raft.


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fn 
1
k
1.8375 104 N m
1

 1.455 Hz  1.5 Hz
2 m 2
220 kg
(b) As explained in the text, for a vertical spring the gravitational potential energy can be ignored if
the displacement is measured from the oscillator’s equilibrium position. The total energy is
thus
Etotal  12 kA2 
1
2
1.8375 10
4

N m 4.0 102 m

2
 14.7 J  15 J .
77. (a) The overtones are given by f n  nf1 , n  2, 3, 4
G : f 2  2  392 Hz   784 Hz
f3  3  392 Hz   1180 Hz
A : f 2  2  440 Hz   880 Hz
f3  3  440 Hz   1320 Hz
(b) If the two strings have the same length, they have the same wavelength. The frequency
difference is then due to a difference in wave speed caused by different masses for the strings.
FT
2
2
mG L
fG vG  vG
mG  f A   440 
mA





  
  1.26
f A vA  vA
mG
mA  f G   392 
FT
mA L
(c) If the two strings have the same mass per unit length and the same tension, then the wave speed
on both strings is the same. The frequency difference is then due to a difference in wavelength.
For the fundamental, the wavelength is twice the length of the string.
fG v  G A 2 LA
LG
f
440




 A 
 1.12
f A v  A G 2 LG
LA
fG 392
(d) If the two strings have the same length, they have the same wavelength. The frequency
difference is then due to a difference in wave speed caused by different tensions for the strings.
FTG
fG
fA

vG 
vA 

vG
vA

m L
FTA

FTG
FTA

FTG
FTA A
2
 f   392 
 G  
  0.794
 f A   440 
2
m L
78. (a) Since the cord is not accelerating to the left or right, the tension in the cord must be the same
everywhere. Thus the tension is the same in the two parts of the cord. The speed difference
will then be due to the different mass densities of the two parts of the cord. Let the symbol 
represent the mass per unit length of each part of the cord.
vH
vL





FT 
FT
H

L
L
H
(b) The wavelength ratio is found as follows.
H  v f H vH



L  v f L vL
L
H
The two frequencies must be the same for the cord to remain continuous at the boundary. If the
two parts of the cord oscillate at different frequencies, the cord cannot stay in one piece,
because the two parts would be out of phase with each other at various times.
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(c) Since H  L , we see that H  L , and so the wavelength is greater in the lighter cord .
79. (a) The maximum speed is given by
vmax  2 f A  2  264 Hz  1.8  10 3 m   3.0 m s .
(b) The maximum acceleration is given by


amax  4 2 f 2 A  4 2  264 Hz  1.8 103 m  5.0 103 m s 2 .
2
80. For the pebble to lose contact with the board means that there is no normal force of the board on the
pebble. If there is no normal force on the pebble, then the only force on the pebble is the force of
gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This
is the maximum downward acceleration that the pebble can have. Thus if the board’s downward
acceleration exceeds g, then the pebble will lose contact. The maximum acceleration and the
amplitude are related by amax  4 2 f 2 A .
amax  4 2 f 2 A  g  A 
g
4 f
2
2

9.8 m s 2
4 1.5 Hz 
2
2
 1.1 101 m
81. For a resonant condition, the free end of the string will be an antinode, and the fixed end of the string
will be a node. The minimum distance from a node to an antinode is  4 . Other wave patterns that
fit the boundary conditions of a node at one end and an antinode at the other end include
3 4 , 5 4 ,
. See the diagrams. The general relationship is L   2n  1  4 , n  1, 2,3, .
Solving for the wavelength gives  
4L
2n  1
, n  1, 2, 3,
.
1
n=1
n=3
0
n=5
0
1
-1
82. The period of a pendulum is given by T  2 L g , and so the length is L 
(a) LAustin 
(b)
LParis 
T 2 g Austin

4 2
T 2 g Paris
4
2

 2.000 s 
2
 9.793m s   0.9922 m
4 2
.
2
4 2
 2.000 s 2  9.809 m
4
T 2g
s2
2
  0.9939 m
LParis  LAustin  0.9939 m  0.9922 m  0.0016 m  1.6 mm
(c) LMoon 
T 2 g Moon
4 2

 2.00 s 
2
1.62 m s   0.164 m
2
4 2
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83. The spring, originally of length l0 , will be stretched downward to a new equilibrium length L when
the mass is hung on it. The amount of downward stretch L  l0 is found from setting the spring force
upward on the mass equal to the weight of the mass: k  L  l0   mg  L  l0  mg k . The length
of the pendulum is then L  l0  mg k . The period of the vertical oscillations is given by
Tver  2 m k , while the period of the pendulum oscillations is given by Tpen  2 L g . Now
compare the periods of the two motions.
Tpen
Tver

2
 l0  mg k 
2 m k
g
Tpen  Tver , by a factor of 1 
84.
l0  mg k

mg k
 1
l0 k
mg
1 
l0 k
mg
Block m stays on top of block M (executing SHM relative to the ground) without slipping due to
static friction. The maximum static frictional force on m is Ffr  s mg . This frictional force causes
max
block m to accelerate, so mamax  s mg  amax  s g . Thus for the blocks to stay in contact
without slipping, the maximum acceleration of block M is also amax   s g . But an object in SHM
has a maximum acceleration given by amax   2 A 
maximum acceleration.
amax 
85.
k
M total
k
M total
A . Equate these two expressions for the
 0.30  9.8 m s2 
s g
A  s g  A 
 M  m 
 6.25 kg   0.14 m
k
130 N m
The speed of the pulses is found from the tension and mass per unit length of the wire.
FT
255 N
v

 143.985m s
mL
 0.123 kg  10.0 m 
The total distance traveled by the two pulses will be the length of the wire. The second pulse has a
shorter time of travel than the first pulse, by 20.0 ms.
L  d1  d 2  vt1  vt2  vt1  v t1  2.00  102


10.0 m   2.00  10 143.985 m s 
 4.4726  102 s
2v
2 143.985 m s 
d1  vt1  143.985 m s   4.4726  102 s   6.44 m
t1 
L  2.00  102 v
2

The two pulses meet 6.44 m from the end where the first pulse originated.
86. For the penny to stay on the block at all times means that there will be a normal force on the penny
from the block, exerted upward. If down is taken to be the positive direction, then the net force on the
penny is Fnet  mg  FN  ma . Solving for the magnitude of the normal force gives FN  mg  ma .
This expression is always positive if the acceleration is upwards (a < 0 ), and so there is no possibility
of the penny losing contact while accelerating upwards. But if a downward acceleration were to be
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larger than g, then the normal force would go to zero, since the normal force cannot switch directions
 FN  0  . Thus the limiting condition is adown  g . This is the maximum value for the acceleration.
2
For SHM, we also know that amax   A 
k
M m
A
k
M
A . Equate these two values for the
acceleration.
k
amax 
A g 
A
Mg
M
k
87. The car on the end of the cable produces tension in the cable, and stretches the cable according to
1 F
Lo , where E is Young’s modulus. Rearrange this equation to see that
Equation (9-4), L 
E A
EA
L , and so the effective spring
the tension force is proportional to the amount of stretch, F 
Lo
constant is k 
EA
. The period of the bouncing can be found from the spring constant and the mass
Lo
on the end of the cable.
m
T  2
k
mLo
 2
1200 kg  22 m 
 2
EA
 200 10
9
 
N m 2  3.2  10 3 m

2
 0.40 s
88. From Equation (9-6) and Figure (9-22c), the restoring force on the top of the Jell-O is F 
GA
L ,
Lo
and is in the opposite direction to the displacement of the top from the equilibrium condition. Thus
GA
the “spring constant” for the restoring force is k 
. If you were to look at a layer of Jell-O
Lo
closer to the base, the displacement would be less, but so would the restoring force in proportion, and
so we estimate all of the Jell-O as having the same spring constant. The frequency of vibration can
be determined from the spring constant and the mass of the Jell-O.
f 

1
k
2
m

1
GA Lo
2
V
1
GA Lo
2
 ALo
520 N m 2
1
2

1300 kg
m3
 4.0 10 m 
2
2

1
G
2
 L2o
 2.5 Hz
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288