Question

A person observes two moving trains. First reaching the station and another leaves the station with equal speed of \(30 \text{ m/s}\). If both trains emit sounds of frequency \(300 \text{ Hz}\), difference of frequencies heard by the person will be (speed of sound in air \(330 \text{ m/s}\))

• A. \(80 \text{ Hz}\)
• B. \(75 \text{ Hz}\)
• C. \(55 \text{ Hz}\)
• D. \(45 \text{ Hz}\)

Hint:

Frequency increases when source approaches and decreases when source recedes.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A stationary observer hears sound from two moving sources (trains).
One source is approaching, and the other is receding at the same speed.
We need to calculate the difference in frequencies heard (beat frequency).
Step 2: Key Formula or Approach:
Doppler effect formula for stationary observer and moving source:
Approaching: \(f_{\text{app}} = f \left( \frac{v}{v - v_s} \right)\)
Receding: \(f_{\text{rec}} = f \left( \frac{v}{v + v_s} \right)\)
Where \(v\) is the speed of sound, \(v_s\) is the source speed, and \(f\) is the actual frequency.
Step 3: Detailed Explanation:
Given: \(f = 300 \text{ Hz}\), \(v = 330 \text{ m/s}\), \(v_s = 30 \text{ m/s}\).
Frequency from the approaching train (\(f_1\)):
\[ f_1 = 300 \left( \frac{330}{330 - 30} \right) = 300 \left( \frac{330}{300} \right) = 330 \text{ Hz} \]
Frequency from the receding train (\(f_2\)):
\[ f_2 = 300 \left( \frac{330}{330 + 30} \right) = 300 \left( \frac{330}{360} \right) = 300 \times \frac{11}{12} = 25 \times 11 = 275 \text{ Hz} \]
Difference in frequencies:
\[ \Delta f = f_1 - f_2 = 330 - 275 = 55 \text{ Hz} \]
Step 4: Final Answer:
The difference of frequencies heard is \(55 \text{ Hz}\).