Question

A source of sound $S$ is moving with a velocity of $50\text{m/s}$ towards a stationary observer. The observer measures the frequency of the source as $1000\text{Hz}$. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air $350\text{m/s})$

• A. $857\text{Hz}$
• B. $1143\text{Hz}$
• C. $750\text{Hz}$
• D. $807\text{Hz}$

Answer 1

The correct option is C $750\text{Hz}$

Given:
Source velocity $V_{source}=50\text{m/s}$
Sound velocity $V_{sound}=350\text{m/s}$
frequency measured by the observer $f_{app}=1000\text{Hz}$
We know that apperent frequency when source is moving towards a stationary observer is given as,

$f_{app}=f_{source}\left(\frac{V_{sound}-V_{observer}}{V_{sound}-V_{source}}\right)$

$1000=f_{source}\left(\frac{350-0}{350-50}\right)$

$f_{source}=1000\times \left(\frac{300}{350}\right)$

In second case when source is moving away from observer

$f^{\prime }=f_{source}\left(\frac{350}{350+50}\right)$

$f^{\prime }=\frac{1000\times 300}{350}\times \frac{350}{400}$

$f^{\prime }\approx 750\text{Hz}$

Hence option $\left(A\right)$ is correct.