Answers to problems for Waves # 35, 38,45, 49 and 51
35.
The wave speed is given by    f . The period is 3.0 seconds, and the wavelength is 8.0 m.
   f   /T  (7.0 m)/(3.0 s)  2.3 m/s
38.
To find the wavelength, use    /f .
AM:
1 

f1

3.00 108 m/s
1 
FM:
45.
(a)
550 103 Hz

f1

 545 m 2 
3.00 108 m/s
88 106 Hz

f2

 3.41 m 2 
3.00 108 m/s
1600 103 Hz

f2

 188 m
3.00 108 m/s
108 106 Hz
AM: 190 m to 550 m
 2.78 m
FM: 2.8 m to 3.4 m
Assuming spherically symmetric waves, the intensity will be inversely proportional to the square of the
distance from the source, as given by Eq. 11–16b. Thus Ir 2 will be constant.
2
2
I near rnear
 I far rfar
I near  I far
49.
2
rfar
2
rnear
 (3.0 106 W/m2 )
(54 km)2
(1.0 km)
2
 8.748 109 W/m 2  8.7 109 W/m 2
The frequencies of the harmonics of a string that is fixed at both ends are given by f n  nf1 , so
the first four harmonics are
51.

f1  440 Hz, f 2  880 Hz, f3  1320 Hz, and f 4  1760 Hz .
Four loops is the standing wave pattern for the fourth harmonic, with a frequency given by
f 4  4 f1  240 Hz.
Thus,
f1  60 Hz, f 2  120 Hz, f3  180 Hz, and f5  300 Hz
are all other resonant
frequencies, where f1 is the fundamental or first harmonic, f 2 is the first overtone or second harmonic,
f3 is the second overtone or third harmonic, and f5 is the fourth overtone or fifth harmonic.