Question

A source of sound emits sound wave of frequency 'f' and moves towards an observer with a velocity $V/3$ where $V$ is the velocity of sound. If the observer moves away from the source with a velocity $V/5$ the apparent frequency heard by him will be ______.

• A. $\frac{15}{2}f$
• B. $\frac{8}{15}f$
• C. $\frac{6}{5}f$
• D. $\frac{15}{18}f$

Hint:

Doppler Sign Convention: "Towards = Higher frequency".
If observer moves towards, add to numerator.
If source moves towards, subtract from denominator.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This is a Doppler Effect problem where both the source and the observer are moving along the same line. The source chases the observer, and the observer runs away from the source.

Step 2: Detailed Explanation:
The general formula for apparent frequency ($f'$) in the Doppler Effect is:
$f' = f \left( \frac{V \pm v_o}{V \mp v_s} \right)$
where:
$f$ = Actual frequency
$V$ = Velocity of sound
$v_o$ = Velocity of observer
$v_s$ = Velocity of source
Let's carefully determine the signs based on the relative motion:
1. Numerator (Observer effect): The observer is moving away from the source. This relative motion tends to decrease the frequency. Therefore, we use a negative sign in the numerator.
$\text{Numerator} = V - v_o$
2. Denominator (Source effect): The source is moving towards the observer. This relative motion tends to increase the frequency. Therefore, we use a negative sign in the denominator (subtracting makes the fraction larger).
$\text{Denominator} = V - v_s$
So the specific formula for this scenario is:
$f' = f \left( \frac{V - v_o}{V - v_s} \right)$
Substitute the given velocities: $v_o = V/5$ and $v_s = V/3$.
$f' = f \left( \frac{V - V/5}{V - V/3} \right)$
Simplify the fractions inside the bracket:
Numerator: $V - V/5 = \frac{5V - 1V}{5} = \frac{4V}{5}$
Denominator: $V - V/3 = \frac{3V - 1V}{3} = \frac{2V}{3}$
Substitute back:
$f' = f \left( \frac{\frac{4V}{5}}{\frac{2V}{3}} \right)$
Multiply by the reciprocal:
$f' = f \left( \frac{4V}{5} \times \frac{3}{2V} \right)$
The $V$ perfectly cancels out:
$f' = f \left( \frac{12}{10} \right)$
$f' = f \left( \frac{6}{5} \right)$

Step 3: Final Answer:
The apparent frequency is $\frac{6}{5}f$, matching option (c).