Name (LAST, First) ________________, ____________
Date ___ / ___ / ______ Block 1 2 3 4 5 6 7 8
Unit 01 – Spring – Waves, Key Chapter 11
Examples:
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.
g
g
9.8 m s 2
amax   2 A  g  A  2  2 2  2
 0.99 m
2

4 f
4  0.50 Hz 
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
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.
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
Questions (Ch 11):
533569032
2/17/2016 5:38:00 AM
Page 1 of 3
Name (LAST, First) ________________, ____________
Date ___ / ___ / ______ Block 1 2 3 4 5 6 7 8
14. From Equation 11-19b, the fundamental frequency of oscillation for a string with both ends fixed is
f1 
v
2L
. The speed of waves on the string is given by Equation 11-13, v 
relationships gives f1 
FT
m L
. Combining these two
FT
1
2
. By wrapping the string with wire, the mass of the string can be greatly
mL
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.
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.
20. The string could be touched at the location of a node
without disturbing the motion, because the nodes do not
move. A string vibrating in three segments has 2 nodes in
addition to the ones at the ends. See the diagram.
1
-2
node
-1
node
Problems (Ch 11):
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
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
 t 
x

FT
m L
28 m
FT
m L
.
 0.35 s
150 N
 0.65 kg   28 m 
43. The speed of the water wave is given by v  B  , where B is the bulk modulus of water, from Table 91, 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
47. (a) Assuming spherically symmetric waves, the intensity will be inversely proportional to the
533569032
2/17/2016 5:38:00 AM
Page 2 of 3
Name (LAST, First) ________________, ____________
Date ___ / ___ / ______ Block 1 2 3 4 5 6 7 8
2
square of the distance from the source. Thus Ir 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
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.
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 reduced to
2L
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
533569032
2/17/2016 5:38:00 AM
Page 3 of 3