A source emits sound waves of frequency `1000 H_(Z)`. The source moves to the right with a speed of `32 m//s` relative to ground. On the right a reflecting surface moves towards left with a speed of `64 m//s` relative to the ground. The speed of sound in air is `332 m//s`. Find
(a) the wavelength of sound in aheed of the source,
(b) the number of waves arriving per second which meets the reflecting surface,
(c ) the speed of reflected waves and
(d) the wavelength of reflected waves.

Explanation

(a) Due to motion of the source the wavelength (and hence the frequency) is actually changed from `lamda` and `lamda'` such that if `n=` actual frequency.
`lamda'=(c-v_S)/(n)=(332-32)/(1000)=0.3m`
(b) The number of waves arriving at the reflecting surface is the same as the number of waves received by an observer moving towards the source. This is given by the apparent frequency.
`n'=(c-v_0)/(c-v_S)xxn`
`=(332-(-64))/(332-32)xx1000=1320Hz`
(c ) Same as that of the incident wave because the speed of wave depends only on the characteristics of the medium. therefore, speed of the reflected wave`=332ms^-1`
(d) Now reflector will act as a source of frequency 1320 Hz. So wavelength emitted by this:
`lamda'=(v-v_(ref))/(1320)=(332-64)/(1320)=0.2m`