Solutions to Problems Set 2
Questions to Ask Yourself
1. What equations am I using for these problems?
- Make a list and make sure you know how to manipulate the equation
for different variables.
2. For which, variables am I trying to solve?
- Make a list of variables – practice naming, units. finding by reading
the problems
3. Practice reading the problem as well as solving the problem.
4. Bring questions to class or post on this discussion page.
36.
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  343 m 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 2  
 188 m
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.
(a) For water:
v B  
(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  103 kg m3
v E  
 4.1 103 m s
200  109 N m 2
7.8  10 kg m
3
3
 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 
v
x
t
FT

m L
x
 t 
28 m

FT
FT
m L
.
 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.
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
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  f n1  f n   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
nv
. The resonant frequencies
, where Lvib is the length
2 Lvib
of the portion that is actually vibrating. Combining these relationships allows the frequencies to be
calculated.
fn 
FT
n
2 Lvib
f1 
m L
f 2  2 f1  581.54 Hz
1
2  0.62 m 
 3.6 10
520 N
3
kg
  0.90 m 
 290.77 Hz
f 3  3 f1  872.31Hz
So the three frequencies are 290 Hz , 580 Hz , 870 Hz , to 2 significant figures.
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 
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
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
1
2
mv12  12 kx12  12 mv22  12 kx22 
1500 kg
v1 
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.
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.
mg
F  kx  mg  x 
 65 kg   9.8 m

s2
  3.2 10
2
m
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.
4
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
x=H
2
2
2
2
mvtop
 mgytop  12 kxtop
 12 mvbottom
 mgybottom  12 kxbottom
0  mgH  0  0  0  12 kH 2 
f 

1

2 2
2g
H

k

m

1
2 9.8 m s
2
0.33 m
2
2g
H
x=0
 2   
y=H
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
  950 kg 
 22 m s 
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  10 4 N m
 0.71 s
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