Question 1
A source of sound wave of frequency 660Hz emits
waves of wavelength 0.12m in air at 40oC. What is the velocity of
sound wave in air at this temperature? What would be the wavelength of sound
wave from the source in air at 70oC?
Answer
Frequency = f = 660Hz
Wavelength= λ = 0.12m
Temperature = 40+273 = 313K
Velocity
of sound wave at 400C
V = fλ = 660 x0.12 =79.2m/s
Wavelength
of wave at 700C
Note: to find the wavelength, we must find the velocity
of the sound wave at 700C. Velocity of sound wave is directly
proportional to the square root of its temperature in kelvin.
V1/V2 = √T1/√T2
V1 = 79.2m/s
T1 = 313K
T2 = 70 + 273 = 343K
79.2/V2 = √313/√343
79.22/(V2)2 =
313/343
313(V2)2 = 79.22
x 343= 2151515.52
(V2)2 = 2151515.52/313
(V2)2 = 6873.9
V2 =
√6873.9 = 82.9m/s
Since V=fλ then
82.9 = 660 x λ
.λ = 82.9/660 =0.13m
Question
II
A sound wave of velocity 200m/s, calculate the temperature
of the wave, if the velocity of sound wave in air is 330m/s at 00C.
Answer
Velocity = V2 = 200m/s
Temperature = T2 =??
Velocity = V1 = 330m/s
Temperature = T1 = 273+0 = 273K
V1/V2 = √T1/√T2
330/200 = √273/ √T2
3302/2002 = 273/T2
T2 = 2002x273/ 3302 = 10920000/108900
= 100.3K
or -1730C
Question
III
The velocity of sound wave in air is 330m/s at 00C.
What would be its velocity if the temperature is increased to 1000C?
What is the change in velocity of the wave?
Answer
V1 = 330m/s
T1 = 0+273 =273K
V2 =??
T2 = 100+273 =373K
Velocity
of wave at 1000C
V1/V2 = √T1 /√T2
330/V2 = √273/ √373
3302/(V2)2 =
273/373
(V2)2 = 330x330x373/273
=
148790.1
V2 = √148790.1 = 385.7m/s
Change
in velocity
V2
- V1 = change in velocity
385.7-330 = 55.7m/s
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