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,........?

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.