Question

A source is moving with constant speed $v_s=20\text{m/s}$ towards a stationary observer due east of the source. Wind is blowing at the speed of $20\text{m/s}$ at ${60}^{\circ }$ north of east. The source has frequency $1000\text{Hz}$. If the speed of sound is $310\text{m/s}$, the frequency observed by the observer is approximately

• A. $1000\text{Hz}$
• B. $1500\text{Hz}$
• C. $1067\text{Hz}$
• D. $1200\text{Hz}$

Answer 1

The correct option is C $1067\text{Hz}$

Given that,
Frequency of source, $f_s=1000\text{Hz}$
Velocity of source, $\left(v_s\right)=20\text{m/s}$
From question

Velocity of wind along source motion $\left(v_w^{\prime }\right)=v_w\cos {60}^{\circ }$
$=20\times \frac{1}{2}=10\text{m/s}$
Velocity of sound $\left(v\right)=310\text{m/s}$
For apparent frequency,
$f_{app}=f_s\left(\frac{v+v_w^{\prime }}{v+v_w^{\prime }-v_s}\right)$
$=1000\left(\frac{310+10}{310+10-20}\right)=1000\times \frac{320}{300}=1067\text{Hz}$

Why this question? Caution:- Be careful when taking the sign convention of wind velocity, source velocity and observer velocity in doppler effect.