Chapter : 5. Sound
Example : 7
A longitudinal wave is produced on a toy string. The wave travels at a speed of 30 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the string ?
Solution. Given, Velocity, v = 30 cm/s
Frequency, ν = 20 Hz
Minimum separation between the two consecutive compressions is equal to one wavelength λ and
Example : 8
Wave of frequency 200 Hz produced in a string is represented in figure. Find out the following :
(i) Amplitude
(ii) Wavelength
(iii) Wave velocity
Solution. (i) Amplitude = Maximum displacement = 10 cm
(ii) Wavelength λ = Distance between two successive crests = 40 cm
(iii) Now, frequency, n = 2 Hz
Wavelength, λ = 40cm = 0.4 m
∴ Wave velocity, v = ν λ = 200 × 0.4 m/s = 80m/s
Example : 9
A stone is dropped into a well 44.1 m deep. The sound of splash is heard 3.13 seconds after the stone is dropped. Calculate the velocity of sound in air.
Solution. First we calculate the time taken by the stone to reach the water level by using the relation:
s = ut +
gt2
Here s = 44.1 m, u = 0, g = 9.8 m/s2
∴ 44.1 = 0 × t +× 9.8 × t2
or
or t = 3 s
Time taken by the sound to reach the top of the well
t2 = 3.13 – 3 = 0.13 s
Now, speed of sound
A longitudinal wave is produced on a toy string. The wave travels at a speed of 30 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the string ?
Solution. Given, Velocity, v = 30 cm/s
Frequency, ν = 20 Hz
Minimum separation between the two consecutive compressions is equal to one wavelength λ and
Example : 8
Wave of frequency 200 Hz produced in a string is represented in figure. Find out the following :
(i) Amplitude
(ii) Wavelength
(iii) Wave velocity
Solution. (i) Amplitude = Maximum displacement = 10 cm
(ii) Wavelength λ = Distance between two successive crests = 40 cm
(iii) Now, frequency, n = 2 Hz
Wavelength, λ = 40cm = 0.4 m
∴ Wave velocity, v = ν λ = 200 × 0.4 m/s = 80m/s
Example : 9
A stone is dropped into a well 44.1 m deep. The sound of splash is heard 3.13 seconds after the stone is dropped. Calculate the velocity of sound in air.
Solution. First we calculate the time taken by the stone to reach the water level by using the relation:
s = ut +
gt2Here s = 44.1 m, u = 0, g = 9.8 m/s2
∴ 44.1 = 0 × t +
× 9.8 × t2or
or t = 3 s
Time taken by the sound to reach the top of the well
t2 = 3.13 – 3 = 0.13 s
Now, speed of sound
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