Question

The frequency of sound heard by an observer moving towards a stationary source with certain speed is $n_1$ and if the observer moves away from the same source with same speed, the frequency of sound heard by the observer is $n_2$. If the speed of sound in air is 340ms$^{-1}$ and $n_1 : n_2 = 71:65$, then speed of observer is

• A. 36 kmph
• B. 27 kmph
• C. 15 kmph
• D. 54 kmph

Hint:

The general Doppler effect formula is $n' = n_0 \left(\frac{v \pm v_o}{v \mp v_s}\right)$. Use the top signs for "towards" motion (frequency increases) and the bottom signs for "away" motion (frequency decreases). For ratio problems, this formula simplifies nicely.

Answer 1

The Correct Option is D

This is a problem involving the Doppler effect for sound.
Let $n_0$ be the actual frequency of the stationary source.
Let $v$ be the speed of sound in air, $v = 340$ m/s.
Let $v_o$ be the speed of the observer.
Case 1: Observer moving towards the stationary source.
The apparent frequency heard is $n_1 = n_0 \left( \frac{v+v_o}{v} \right)$.
Case 2: Observer moving away from the stationary source.
The apparent frequency heard is $n_2 = n_0 \left( \frac{v-v_o}{v} \right)$.
We are given the ratio $n_1 : n_2 = 71:65$.
$\frac{n_1}{n_2} = \frac{n_0 \left( \frac{v+v_o}{v} \right)}{n_0 \left( \frac{v-v_o}{v} \right)} = \frac{v+v_o}{v-v_o}$.
So, $\frac{v+v_o}{v-v_o} = \frac{71}{65}$.
Cross-multiply to solve for $v_o$:
$65(v+v_o) = 71(v-v_o)$.
$65v + 65v_o = 71v - 71v_o$.
$65v_o + 71v_o = 71v - 65v$.
$136v_o = 6v$.
$v_o = \frac{6v}{136} = \frac{3v}{68}$.
Substitute the value of $v = 340$ m/s:
$v_o = \frac{3 \times 340}{68} = 3 \times 5 = 15$ m/s.
The options are in kmph. We need to convert the speed.
To convert m/s to kmph, multiply by $\frac{18}{5}$.
Speed of observer = $15 \text{ m/s} \times \frac{18}{5} = 3 \times 18 = 54$ kmph.