A source of sound with frequency $256\text{ Hz}$ is moving with velocity $v$ towards a wall. When the…
A source of sound with frequency $256\text{ Hz}$ is moving with velocity $v$ towards a wall. When the observer is between source and the wall, he finds that the frequency of two waves received directly from the source is $x$ and the frequency of the waves received after reflection from the wall is $y$, then
$x > y$
$x < y$
$x = y$
Nothing can be said
Solution
Both S and S' are moving towards observer. So, both the observed frequencies will be more than the actual frequency but both will be equal.
[Diagram showing source S moving right at speed $v_s$ towards observer O, and image source S' moving left at speed $v_s$.]