Question

The apparent frequency of a note, when a listener moves towards a stationary source, with velocity of 40 m/s is 200 Hz. When he moves away from the same source with the same speed, the apparent frequency of the same note is 160 Hz. The velocity of sound in air is : (in m/s)

• A. 360
• B. 330
• C. 320
• D. 340

Answer 1

Answer: (A)

360

Let the original frequency of the source be $f_o$.
Doppler effect when an observer moves towards the stationary source :
Apparent frequency heard $f^{\prime }=f_o[\frac{v_{sound}+v_o}{v_{sound}}]$
$\therefore$ $200=f_o[\frac{v_{sound}+40}{v_{sound}}]$ .....(1)
Doppler effect when an observer moves away from the stationary source :
Apparent frequency heard $f^{\prime \prime }=f_o[\frac{v_{sound}-v_o}{v_{sound}}]$
$\therefore$ $160=f_o[\frac{v_{sound}-40}{v_{sound}}]$ .....(2)
Dividing (1) and (2), we get
$\frac{200}{160}=\frac{v_{sound}+40}{v_{sound}-40}$
$5v_{sound}-200=4v_{sound}+160$
$⟹$ $v_{sound}=360m/s$