Question

A $600\text{Hz}$ source, an observer and the wind are moving as shown in the figure with respect to the ground. Find the frequency heard by the observer. Take sound speed in stationary air to be $330\text{m/s}$.

• A. $500\text{Hz}$
• B. $550\text{Hz}$
• C. $525\text{Hz}$
• D. $425\text{Hz}$

Answer 1

The correct option is C $525\text{Hz}$

Given,
velocity of observer $\left(v_o\right)=30\text{m/s}$
velocity of source $\left(v_s\right)=12\text{m/s}$

velocity of sound in stationary air $\left(v\right)=330\text{m/s}$

velocity of wind $\left(v_w\right)=8\text{m/s}$

When the source and observer are moving away from each other in a stationary medium, we can write,
$f_{app}=f_0\left(\frac{v-v_o}{v+v_s}\right)$

When the medium is not stationary and velocity of medium is $v_w$, actual velocity of sound with respect to the ground is given by $v\pm v_w$

Thus, we can write that,

$f_{app}=f_0\left(\frac{v\pm v_w-v_o}{v\pm v_w+v_s}\right)$

From the diagram, we can deduce that the direction of wind is from the observer to the source.

So, $f_{app}=f_0\left(\frac{v-v_w-v_o}{v-v_w+v_s}\right)$

From the data given in the question,

$f_{app}=600\left(\frac{330-8-30}{330-8+12}\right)=\frac{600\times 292}{334}\approx 525\text{Hz}$