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: Formula
Apparent frequency $n' = n \left( \frac{v}{v \mp v_s} \right)$.

Step 2: Approaching Train ($n_1$)
$n_1 = 300 \left( \frac{330}{330 - 30} \right) = 300 \left( \frac{330}{300} \right) = 330 \text{ Hz}$.

Step 3: Receding Train ($n_2$)
$n_2 = 300 \left( \frac{330}{330 + 30} \right) = 300 \left( \frac{330}{360} \right) = 275 \text{ Hz}$.

Step 4: Difference
$\Delta n = n_1 - n_2 = 330 - 275 = 55 \text{ Hz}$.
Final Answer: (C)