Question

A person $B$ accidentally slips from a large building and screams with a sound of constant frequency $n$ as he falls as shown in the figure. The apparent frequency of the sound of the scream as heard by the person $A$, at the top of the building is dependent on time as ($v$ is speed of sound)

• A. $n' = n \left(\frac{ v }{ v - gt }\right)$
• B. $n ' = n \left(\frac{ v }{ v + gt }\right)$
• C. $n ' = n \left(\frac{2 v }{2 v + gt ^{2}}\right)$
• D. $n' = n \left(\frac{2 v }{2 v - gt ^{2}}\right)$

Answer 1

Answer: (B)

The person $B$ moves away from person $A$
or the source of sound moves away from $A$.
Hence, the apparent frequency is
$n'= n \left(\frac{ v }{ v + v _{ s }}\right)$
where $v _{ s }$ is the speed of sound and $v _{ B }$ is the speed of the person B(source)
Since $($ from $v = u + at$ and $u =0$ )
$v _{ S }= gt$
we get, $n'= n \left(\frac{ v }{ v + gt }\right)$