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. ${\left(\nu \right)}^{{}^{\prime }}=\nu \left(\frac{v}{v-gt}\right)$
• B. ${\left(\nu \right)}^{{}^{\prime }}=\nu \left(\frac{v}{v+gt}\right)$
• C. ${\left(\nu \right)}^{{}^{\prime }}=\nu \left(\frac{2v}{2v+g{t}^2}\right)$
• D. ${\left(\nu \right)}^{{}^{\prime }}=\nu \left(\frac{2v}{2v-g{t}^2}\right)$

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