Question

A source of sound with frequency 256 Hz is moving with a velocity $v$ towards a wall and an observer is stationary between the source and the wall. When the observer is between source and the wall, he finds that the frequency of the waves received directly from the source as $x$ and the frequency of the waves received after reflection from the wall as $y$. Then :

• A. $x>y$
• B. $x<y$
• C. $x=y$
• D. cannot be predicted

Answer 1

The correct option is C $x=y$

Let $x$ and $y$ be the frequencies heard by the observer O coming

directly from the source and after reflecting from the wall respectively.

Velocity of sound in air be $v_{sound}m/s$

Velocity of the source $v_{source}=vm/s$

Doppler effect when a source moves towards the stationary observer:

$x=f[\frac{v_{sound}}{v_{sound}-v_{source}}]$ $=256[\frac{v_{sound}}{v_{sound}-v}]$

Doppler effect when a source moves towards the wall:

$n^{\prime }=f[\frac{v_{sound}}{v_{sound}-v_{source}}]$ $=256[\frac{v_{sound}}{v_{sound}-v}]$

Now apparent frequency after the reflection from the stationary wall (no Doppler effect) $y=n^{\prime }=256[\frac{v_{sound}}{v_{sound}-v}]$

$⟹x=y$ $=256[\frac{v_{sound}}{v_{sound}-v}]$