Question

An observer moves towards a stationary source of sound, with a velocity ${\frac{1}{5}}^{th}$ of the velocity of sound. What is the $\%$ increase in the apparent frequency?

• A. Zero
• B. $0.5\%$
• C. $5\%$
• D. $20\%$

Answer 1

The correct option is D

$20\%$

The explanation for the correct option

Step 1: Given data

An observer moves toward a stationary source of the sound, with a velocity one-fifth of the velocity of sound

$\therefore v_0=\frac{v}{5}$

Step 2: Formula used

$f\text{'}=f\left(\frac{v+v_o}{v-v_s}\right)$ $v_s$ is the velocity of the source of the sound, $v_o$ is the velocity of the observer $v$ is the velocity of the sound$f\text{'}$ is the apparent frequency and $f$ is the frequency

Step 3: Find the apparent frequency

According to the doppler effect theory, the perceived frequency heard by the observer increases as they get closer to a stationary source of the sound. In this case, the apparent frequency heard is as follows:

$\therefore v_0=\frac{v}{5}$

As we see the source is stationary, so $v_s=0$

It is given $v_0=\frac{v}{5}$

$$\begin{gathered} f\text{'}=f\left(\frac{v+v_o}{v-v_s}\right) \\ =f\left(\frac{v+{\displaystyle \frac{v}{5}}}{v}\right) \\ =1.2f \end{gathered}$$

Step 4: Find the increase in apparent frequency

The change in apparent frequency is, $=\frac{f\text{'}-f}{f}\times 100$

$$\begin{gathered} =\frac{1.2f-f}{f}\times 100 \\ =20\% \end{gathered}$$

Hence option D is correct