Question:
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 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
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One train approaches and one recedes, so use the Doppler formula for each.
Updated On: Oct 1, 2026
- \(10 \text{Hz}\)
- \(33 \text{Hz}\)
- \(55 \text{Hz}\)
- \(80 \text{Hz}\)
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The Correct Option is C
Solution and Explanation
Step 1: Approaching train
\(f_A = f\frac{v}{v - v_s} = 300\times\frac{330}{300} = 330\ \text{Hz}\).
Step 2: Receding train
\(f_B = f\frac{v}{v + v_s} = 300\times\frac{330}{360} = 275\ \text{Hz}\).
Step 3: Difference
\(330 - 275 = 55\ \text{Hz}\). Option (C).
Final Answer:
The difference is 55 Hz. \[ \boxed{\text{(C)}\ 55\ \text{Hz}} \]
\(f_A = f\frac{v}{v - v_s} = 300\times\frac{330}{300} = 330\ \text{Hz}\).
Step 2: Receding train
\(f_B = f\frac{v}{v + v_s} = 300\times\frac{330}{360} = 275\ \text{Hz}\).
Step 3: Difference
\(330 - 275 = 55\ \text{Hz}\). Option (C).
Final Answer:
The difference is 55 Hz. \[ \boxed{\text{(C)}\ 55\ \text{Hz}} \]
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