Q171 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
The source is emitting sound of frequency 'n'. A source and a listener both are moving towards each other with a speed equal to $\left(\dfrac{V}{3}\right)$, where V is the speed of sound. The frequency heard by the listener will be
- A$\dfrac{n}{3}$
- B$\dfrac{n}{2}$
- Correct answerC$2n$
- D$3n$
Solution
The apparent frequency heard by the listener is given by the Doppler effect formula:
$n' = n \left( \dfrac{V + v_L}{V - v_S} \right)$
Since both the source and the listener are moving towards each other with speed $\dfrac{V}{3}$:
$v_L = \dfrac{V}{3}$
$v_S = \dfrac{V}{3}$
Substituting these values into the formula:
$n' = n \left( \dfrac{V + \dfrac{V}{3}}{V - \dfrac{V}{3}} \right)$
$n' = n \left( \dfrac{\dfrac{4V}{3}}{\dfrac{2V}{3}} \right)$
$n' = 2n$
Answer: $2n$
$n' = n \left( \dfrac{V + v_L}{V - v_S} \right)$
Since both the source and the listener are moving towards each other with speed $\dfrac{V}{3}$:
$v_L = \dfrac{V}{3}$
$v_S = \dfrac{V}{3}$
Substituting these values into the formula:
$n' = n \left( \dfrac{V + \dfrac{V}{3}}{V - \dfrac{V}{3}} \right)$
$n' = n \left( \dfrac{\dfrac{4V}{3}}{\dfrac{2V}{3}} \right)$
$n' = 2n$
Answer: $2n$
Hence, the correct answer is option: 'C'.