Question

A person $B$ accidentally slips from a large building and screams with a sound of constant frequency $\nu$ as he falls, as shown in the figure. What will be the apparent frequency of the sound of the scream as heard by the person $A$ , at the top of the building as a function of time?

• A. $\left(\nu\right)^{'}=\nu\left(\frac{v}{v - g t}\right)$
• B. $\left(\nu\right)^{'}=\nu\left(\frac{v}{v + g t}\right)$
• C. $\left(\nu\right)^{'}=\nu\left(\frac{2 v}{2 v + g t^{2}}\right)$
• D. $\left(\nu\right)^{'}=\nu\left(\frac{2 v}{2 v - g t^{2}}\right)$

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
$\left(\nu\right)^{'}=\nu\left(\frac{v}{v + v_{B}}\right)$
where $v$ is the speed of sound and $v_{B}$ is the speed of the person $B\left(\right.\text{ source }\left.\right)$
since (from $v_{B}=u+at$ and $u=0$ )
$v_{B}=gt$
we get, $\left(\nu\right)^{'}=\nu\left(\frac{v}{v + gt}\right)$ .