Physics 103
Spring 2008
Homework 2 Solutions
Giancoli Chapter 11
Questions:
13. The speed of the transverse wave is measuring how fast the wave disturbance moves along the cord.
For a uniform cord, that speed is constant, and depends on the tension in the cord and the mass
density of the cord. The speed of a tiny piece of the cord is measuring how fast the piece of cord
moves perpendicularly to the cord, as the disturbance passes by. That speed is not constant — if a
sinusoidal wave is traveling on the cord, the speed of each piece of the cord will be given by the
speed relationship of a simple harmonic oscillator (Equation 11-9), which depends on the amplitude
of the wave, the frequency of the wave, and the specific time of observation.
14. From Equation 11-19b, the fundamental frequency of oscillation for a string with both ends fixed is
f1 
v
. The speed of waves on the string is given by Equation 11-13, v 
2L
two relationships gives f1 
1
2
FT
. Combining these
m L
FT
. By wrapping the string with wire, the mass of the string can be
mL
greatly increased without changing the length or the tension of the string, and thus the string has a
low fundamental frequency.
15. If you strike the horizontal rod vertically, you will create primarily transverse waves. If you strike
the rod parallel to its length, you will create primarily longitudinal waves.
17. (a)
Similar to the discussion in section 11-9 for spherical waves, as a circular wave expands, the
circumference of the wave increases. For the energy in the wave to be conserved, as the
circumference increases, the intensity has to decrease. The intensity of the wave is proportional
to the square of the amplitude
(b)
The water waves will decrease in amplitude due to dissipation of energy from viscosity in the
water (dissipative or frictional energy loss).
19. The frequency must stay the same because the media is continuous — the end of one section of
cord is physically tied to the other section of cord. If the end of the first section of cord is
vibrating up and down with a given frequency, then since it is attached to the other section of
cord, the other section must vibrate at the same frequency. If the two pieces of cord did not
move at the same frequency, they would not stay connected, and then the waves would not pass
from one section to another.
Chapter 11 Problems
52. The frequencies of the harmonics of a string that is fixed at both ends are given by f n  nf1 , and so
f1  440 Hz, f 2  880 Hz, f 3  1320 Hz, f 4  1760 Hz .
the first four harmonics are
55. Adjacent nodes are separated by a half-wavelength, as examination of Figure 11-40 will show.

v
f

xnode  12  
v
92 m s

 9.7  10 2 m
2 f 2 475 Hz 
58. From Equation (11-19b), fn  nv , we see that the frequency is proportional to the wave speed on the
2L
stretched string. From equation (11-13), v 
FT
, we see that the wave speed is proportional to the
m L
square root of the tension. Thus the frequency is proportional to the square root of the tension.
FT2
f
 2
FT1
f1
2

2
f 
 200 Hz 
FT2   2  FT1  
F  0.952 FT1
 205 Hz  T1
 f1 
Thus the tension should be decreased by 4.8% .
60. The tension in the string is the weight of the hanging mass, FT  mg. The speed of waves on the
string can be found by v 
FT

m L
mg and the frequency is given as f  60 Hz. The wavelength
,
m L 
of waves created on the string will thus be given by
v 1
 
f
f
mg
1

m L  60 Hz
0.080 kg 9.80
3.9  10
4
m s2
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, L , 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.
Chapter 12 Questions
3.
The child speaking into a cup creates sound waves which cause the bottom of the cup to vibrate.
Since the string is tightly attached to the bottom of the cup, the vibrations of the cup are transmitted
to longitudinal waves in the string. These longitudinal waves travel down the string, and cause the
bottom of the receiver cup to vibrate. This relatively large vibrating surface moves the adjacent air,
and generates sound waves from the bottom of the cup, traveling up into the cup. These waves are
incident on the receiver’s ear, and they hear the sound from the speaker.
5.
Listening to music while seated far away from the source of sound gives evidence that the speed of
sound in air does not depend on frequency. If the speed were highly frequency dependent, then high
and low sounds created at the same time at the source would arrive at your location at different
times, and the music would sound very disjointed. The fact that the music “stays together” is
evidence that the speed is independent of frequency.
7.
The basic equation determining the pitch of the organ pipe is either f closed 
closed pipe, or fopen 
nv
, n  odd integer, for a
4L
nv
, n  integer, for an open pipe. In each case, the frequency is proportional to
2L
the speed of sound in air. Since the speed is a function of temperature, and the length of any
particular pipe is fixed, the frequency is also a function of temperature. Thus when the temperature
changes, the resonant frequencies of the organ pipes change as well. Since the speed of sound
increases with temperature, as the temperature increases, the pitch of the pipes increases as well.
Chapter 12 Problems
4.
(a)
For the fish, the speed of sound in seawater must be used.
d  vt  t 
(b)
d 1.0 10 3 m

 0.64 s
v
1560 m s
For the fishermen, the speed of sound in air must be used.
d  vt  t 
d 1.0 10 3 m

 2.9 s
v
343 m s
6.
The two sound waves travel the same distance. The sound will travel faster in the concrete, and thus
take a shorter time.
d  vair t air  vconcretet concrete  vconcrete t air  1.1 s   t air 
vconcrete
1.1 s
vconcrete  vair
 vconcrete

d  vair t air  vair 
1.1 s 
 vconcrete  vair

The speed of sound in concrete is obtained from Equation (11-14a), Table (9-1), and Table (10-1).
E
vconcrete 

20  10 9 N m 2
 2949 m s
2.3  10 3 kg m 3



2949 m s
d  vair t air  343 m s 
1.1 s  427 m  4.3  10 2 m
 2949 m s  343 m s

8.
120 dB  10 log
20 dB  10 log
I120
I0
I 20
I0




I120  1012 I 0  1012 1.0  10 12 W m 2  1.0 W m 2


I 20  10 2 I 0  10 2 1.0  10 12 W m 2  1.0  10 10 W m 2
The pain level is 10 10 times more intense than the whisper.
17. The intensity is proportional to the square of the amplitude.
2.0 dB  10 log
I 2.0
A2
A
 10 log 2.02  20 log 2.0
I0
A0
A0

A2.0
 10 0.1  1.259  1.3
A0