(a) Explain how ‘echo’ is produced.
(b) What should be the minimum distance between the listener and the reflector to hear an echo of sound propagating with a speed v m s–1?
(c) Does the speed of sound increase or decrease on a hotter day? Justify.
OR
Establish the relationship between speed of sound, its wavelength and frequency. If speed of sound in air is 340 m s–1, calculate.
(a) wavelength when frequency is 256 Hz.
(b) frequency when wavelength is 0.85 m.
(a) Echo is a reflection of sound that arrives at the listener with a delay after the direct sound. The delay is proportional to the distance of the reflecting surface from the source and the listener. To hear the distinct echo the time interval between the original sound and reflected one must be at least 0.1s
(b) Time interval between the original sound and reflected = 0.1s
Speed = v ms–1
Distance = 0.1 v (total distance covered by the sound from the point of generation to the reflecting surface and back)
Thus, the minimum distance between the listener and the reflector to hear an echo = 1/2 (0.1v)
= 0.05v m
(c) The speed of sound depends on temperature, as the temperature increases the speed of sound also increases. Is the source and the reflector are kept at the minimum distance of echo, the reflected sound will reach early on a hotter day. And it will be difficult to recognize the echo. Whereas if source and reflector are at sufficient distance echo can be heard clearly.
OR
(i) The speed of sound is defined as the distance which a point on a wave such as a compression or a rarefaction travels per unit time.
Speed v = distance/time
= λ/T
Here is the wavelength of the sound wave. It is the distance travelled by the sound wave in one-time period T of the wave. Thus,
ϑ = v (1/T = v)
i.e. speed = wavelength × frequency
The speed of sound remains almost the same for all frequencies in a given medium under the same physical conditions.
(ii) (a) wavelength = speed/frequency
= 340 ms-1/256Hz
= 1.328 m
(b) Frequency = speed/ wavelength
= 340ms-1/0.85 Hz
= 400 Hz