Q125 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
A person observes two moving trains. 'A' reaching the station and 'B' leaving the station with equal speed of 30m/s. If both trains emit sounds with frequency 300 Hz, (speed of sound = 330 m/s) difference of frequencies heard by the person will be
- A$10\ \text{Hz}$
- B$33\ \text{Hz}$
- Correct answerC$55\ \text{Hz}$
- D$80\ \text{Hz}$
Solution
Given:
Speed of sound, $v = 330\ \text{m/s}$
Speed of trains, $v_s = 30\ \text{m/s}$
Actual frequency, $f_0 = 300\ \text{Hz}$
Apparent frequency of train A (moving towards the observer):
$f_A = f_0 \left( \dfrac{v}{v - v_s} \right)$
$f_A = 300 \left( \dfrac{330}{330 - 30} \right) = 300 \left( \dfrac{330}{300} \right) = 330\ \text{Hz}$
Apparent frequency of train B (moving away from the observer):
$f_B = f_0 \left( \dfrac{v}{v + v_s} \right)$
$f_B = 300 \left( \dfrac{330}{330 + 30} \right) = 300 \left( \dfrac{330}{360} \right) = 275\ \text{Hz}$
Difference in frequencies heard by the person:
$\Delta f = f_A - f_B = 330 - 275 = 55\ \text{Hz}$
Answer: $55\ \text{Hz}$
Speed of sound, $v = 330\ \text{m/s}$
Speed of trains, $v_s = 30\ \text{m/s}$
Actual frequency, $f_0 = 300\ \text{Hz}$
Apparent frequency of train A (moving towards the observer):
$f_A = f_0 \left( \dfrac{v}{v - v_s} \right)$
$f_A = 300 \left( \dfrac{330}{330 - 30} \right) = 300 \left( \dfrac{330}{300} \right) = 330\ \text{Hz}$
Apparent frequency of train B (moving away from the observer):
$f_B = f_0 \left( \dfrac{v}{v + v_s} \right)$
$f_B = 300 \left( \dfrac{330}{330 + 30} \right) = 300 \left( \dfrac{330}{360} \right) = 275\ \text{Hz}$
Difference in frequencies heard by the person:
$\Delta f = f_A - f_B = 330 - 275 = 55\ \text{Hz}$
Answer: $55\ \text{Hz}$
Hence, the correct answer is option: 'C'.