2130 PHY Homework solutions Assignment 13
14.4
At T = 27C = 300 K , the speed of sound in air is v = ( 331 m s ) T = ( 331 m s ) 273 K 300 K = 347 m s 273 K
The wavelength of the 20 Hz sound is = 20 000 Hz is =
v 347 m s = = 17 m , and that of the f 20 Hz
347 m s = 1.7 10-2 m=1.7 cm . Thus, range of wavelengths of 20 000 Hz
1.7 cm to 17 m
audible sounds at 27C is 14.25
.
With the train approaching at speed v t , the observed frequency is 345 m s + 0 345 m s 442 Hz = f = f 345 m s - v t 345 m s - v t As the train recedes, the observed frequency is 345 m s + 0 441 H z = f = 345 m s - ( -v t ) Dividing equation (1) by (2) gives 345 m s f 345 m s + v t (2) (1)
442 345 m s + v t = , 441 345 m s - v t
0.391 m s
and solving for the speed of the train yields v t =
14.36
The mass per unit length of the wire is
=
m 0.300 103 kg = = 4.29 kg 104 m , 2 L 70.0 10 m
and the speed of transverse waves is v= F 600 N = = 1.18 103 m s 4.29 104 kg m
The fundamental or first harmonic of the wire has a wavelength of 1 = 2L = 1.40 m , and v 1.18 103 m s f1 = = = 845 H z . frequency 1 1.40 m The frequency of the second harmonic is f 2 = 2 f1 = harmonic is f 3 = 3 f1 =
2.54 10 H z
3
1.69 3 H z 10
and that of the third
.
14.45
Hearing would be best at the fundamental resonance, so = 4L = 4 ( 2.8 cm ) and f= 340 m s 100 cm v 3 = = 3.0 10 Hz = 4 ( 2.8 cm ) 1 m
3.0 kH z
14.47
(a) The speed of sound is 331 m s at 0 C, so the fundamental wavelength of the pipe open at both ends is
1 = 2 L =
v 331 m s v = = giving L = f1 2 f1 2 ( 300 Hz )
0.552 m
(b) At T = 30 C = 303 K , v = ( 331 m s ) f1 =
T 303 K = ( 331 m s ) = 349 m s 273 K 273 K
316 H z
and
349 m s v v = = = 1 2 L 2 ( 0.552 m )