Question

When a source of sound of frequency 600 Hz moves with velocity 30 m/s towards a stationary observe, the apparent wavelength is λ. If the same source is kept stationary and the observer is moved with same velocity 30 m/s towards the source,........?

then apparent wavelength is λ', then find λ' – λ (in cm) for speed of sound in air to be 340 m/s.

Answer 1

Based on the calculations below, the difference in wavelength between these two situations is 10 cm.

Explanation:

First of all, let's determine the wavelength of the sound signal emitted by the stationary source.

This is `lambda=v/f_s` = `340/600 = 0.57` m or 57 cm

The wavelength observed by a stationary listener when a sound source moves toward him is

`lambda=(v/f_s)-v_sT`

where `f_s` is the actual frequency being emitted by the source (600 Hz in this problem), and `v_s` is the speed of the source (30 m/s) and `v` is the speed of sound.

The first term is the unaffected wavelength for a stationary source. The second term is the amount the source moves forward in one period `T` of the sound. Since `T=1/f_s`

This allows us to write

`lambda=(v/f_s)-(v_s/f_s)=(v-v_s)/f_s`, the apparent wavelength for the moving source.

The situation is different for a moving listener. This time, we write

`lambda'=(v/f_s)-v_lT`

Where `v_l` is the speed at which the listener travels. Once again, substituting for `T` allows us to write

`lambda'=(v/f_s)-(v_l/f_s) = (v-v_l)/f_s`

To evaluate `lambda - lambda'` all we do is subtract the above results

`lambda - lambda'=((v-v_s)/f_s)-((v-v_l)/f_s)`

The result is

`lambda - lambda'=(v_l+v_s)/f_s`

Putting in the numbers, we get

`lambda - lambda'=(30+30)/600=60/600 = 0.1 m` or 10 cm.