Question

A source of sound is travelling with a velocity of $30\text{m/s}$ towards a stationary observer. If actual frequency of the source is $800\text{Hz}$ and the wind is blowing with velocity $10\text{m/s}$ in a direction making an angle ${60}^{\circ }$ with the direction of motion of the source, then the apparent frequency heard by the observer is:
(speed of sound in still air is $340\text{m/s}$)

• A. $800\text{Hz}$
• B. $825\text{Hz}$
• C. $876\text{Hz}$
• D. $1000\text{Hz}$

Answer 1

The correct option is C $876\text{Hz}$

Given that
velocity of source $v_s=30\text{m/s}$
velocity of observer $v_o=0\text{m/s}$
velocity of sound in still air $v=340\text{m/s}$

Speed of wind $\left(v_w\right)=10\text{m/s}$

Frequency of sound when the source and observer are initially at rest $\left(f_0\right)=800\text{Hz}$

Given that, wind is blowing in the direction making ${60}^{\circ }$ with the direction of motion of source.

$\therefore$ Speed of wind along the direction of source is $v_w^{\prime }=10\cos {60}^{\circ }$

So, from formula,

$f_{app}=f_0\left(\frac{v\pm v_w\pm v_0}{v\pm v_w\pm v_s}\right)$

Since wind is blowing from source to observer, source is moving towards observer and observer is stationary,
$f_{app}=f_0\left(\frac{v+v_w^{\prime }}{v+v_w^{\prime }-v_s}\right)$

From the data given in the question,

$f_{app}=800\times \left(\frac{340+10\cos {60}^{\circ }}{340+10\cos {60}^{\circ }-30}\right)$

$f_{app}=876.19\approx 876\text{Hz}$