Question

The frequency changes by $20\%$ as a source approaches a stationary observer with constant speed $v_s$. What would be the percentage change in frequency as the source recedes from the observer with the same speed? [Given, $v_s<<v$(velocity of sound)]

• A. $14.3\%$
• B. $16.7\%$
• C. $30\%$
• D. $10\%$

Answer 1

The correct option is A $14.3\%$

Let $f_0$ be the frequency of sound heard by the observer, when both the source of sound and observer are at rest.

Let $v$ be the velocity of sound in the stationary medium, $v_s$ be the velocity of the source of sound and $v_o$ be the velocity of the observer.

Given,
Velocity of observer $\left(v_o\right)=0\text{m/s}$

Case -I:
When the source approaches the observer, the frequency heard by the observer is given by

$f_{app}=f_o\left(\frac{v}{v-v_s}\right)=f_o\left(\frac{1}{1-\frac{v_s}{v}}\right)$

$\Rightarrow \frac{f_{app}}{f_0}=\frac{1}{1-\frac{v_s}{v}}......\left(1\right)$

We can rewrite the above equation as,

$\left(\frac{f_{app}-f_0}{f_0}\right)\times 100=\frac{\frac{v_s}{v}}{1-\frac{v_s}{v}}\times 100$

From the data given in the question,

$\left(\frac{f_{app}-f_0}{f_0}\right)\times 100=20$

$\Rightarrow v=0.2\times (1-\frac{v_s}{v})$

$\Rightarrow v=6v_s.......\left(2\right)$

Using this in $\left(1\right)$, we get

$\frac{f_{app}}{f_0}=1.2$

Case -II:
When the source recedes from the observer at same constant velocity $v_s$,

$f_{app}^{\prime }=f_0\left(\frac{v}{v+v_s}\right)=f_0\left(\frac{1}{1+\frac{v_s}{v}}\right)$

$\Rightarrow \frac{f_{app}^{\prime }}{f_0}=\frac{1}{1+\frac{v_s}{v}}..........\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$, we get

$\frac{f_{app}^{\prime }}{f_0}=\frac{1}{1+\frac{1}{6}}=\frac{6}{7}.......\left(4\right)$

So,

$\left(\frac{f_{app}^{\prime }-f_0}{f_0}\right)\times 100=-0.143\times 100=-14.3\%$

Therefore, observed frequency as the source recedes decreases by $14.3\%$