Question

A source of sound $S$ moving with constant speed approaches a junction $A$. If the original frequency produced by the source is $640\text{Hz}$ then for the situation shown in the figure, the apparent frequency heard by the observer at rest at $O$ will be
[speed of sound $=340\text{m/s}$]

• A. $720\text{Hz}$
• B. $660\text{Hz}$
• C. $700\text{Hz}$
• D. $680\text{Hz}$

Answer 1

The correct option is D $680\text{Hz}$

From Doppler's effect, the apparent frequency heard by the observer is,
$f^{{}^{\prime }}=f_0\left(\frac{v\pm v_0}{v\pm v_s}\right)$, here,
$f_0$ is the original frequency produced by the source, $v$ is the speed of sound, $v_0$ is the speed of observer and $v_s$ is the speed of source.
The frequency heard by observer,
$f^{{}^{\prime }}=f_0\left(\frac{v+v_0}{v-v_s}\right)$


$\Rightarrow f^{{}^{\prime }}=f_0\left(\frac{v+v_0}{v-v_s\cos \theta }\right)$
[we take speed of source and observer along line joining the source and observer]
$\Rightarrow f^{{}^{\prime }}=640\times \left(\frac{340+0}{340-\frac{100}{3}\times \frac{3}{5}}\right)$
$\Rightarrow f^{{}^{\prime }}=680\text{Hz}$