Physics 198
Homework 7
Due September 17
1. (a) Calculate wavelengths in air for sound in the maximum range of human hearing, 20 Hz to
20,000 Hz. (b) What is the wavelength of a 10-MHz ultrasonic wave? Assume that speed of
sound is 343 m/s.
Solution:
(a)
(b)
20 Hz 
v

343m s
 17 m
20 kHz 
v

343m s
f
20 Hz
f 2.0 10 Hz
So the range is from 1.7 cm to 17 m.
v
343m s
10 MHz  
 3.4 105 m=34  m
f 10 106 Hz
4
 1.7  10 2 m
2. A 3.00 m long string with the linear mass density 0.150 g/m is connected to a vibrator with
frequency 120 Hz. The string is oscillating in its 6 th harmonic. Find a) the wave length of this
standing wave; b) the wave length of the fundamental mode; c) the frequency of the
fundamental mode; d) the speed waves in the string; e) the tension in the string. f) What mass
should be attached to this string (as in the lab) to create this tension? g) What tension should be
applied to this string to observe third harmonic if vibrator is the same?
Solution:
2L
2  3.00m
 6 
 1.00m
a) n 
n
6

2L
 2  3.00m  6.00m . (Another solution: n  1  1  nn  6  1.00m  6.00m )
b) 1 
1
n
c) f n  nf1  f1  f n n  120 Hz  6  20 Hz
d) v  1 f1  6 f 6  1.00m120Hz   120m / s
1kg 
2
  2.16 N
 F  v 2   120m / s   0.150 g / m
1000 g 

2.16 N
 0.22kg
f) m  F / g 
9.8m / s 2
e) v 
F
μ
2
2
F 6
 2L 
g) F  v 2   f 2   
f    3     4  F3  4 F6  4  2.16 N  8.64 N
F6  3 
 n 
Another (long) solution:
2
2
1kg 
 2L 
 2  3.00m
 
2
2
  8.64 N
F  v   f    
f   
120 Hz   0.150 g / m
3
1000 g 
 n 

 
Physics 198
Homework 7
Due September 17
3. The fundamental frequency of a string is 440Hz. The length of this string is 32.0 cm, and the
linear mass density is 0.63 g/m. a) Find the frequencies of the first two subharmonics.
b) What is the tension in this string?
Solution:
a) f n  nf1  f 2  2 f1  880Hz , f 3  3 f1  1320Hz
b)
v
F
μ
1kg 
2
2
2
  50 N
 F  v 2   f    2 Lf1    2  0.320m 440 Hz   0.630 g / m
1000 g 
