Question

Two trains \(A\) and \(B\) are moving towards each other with speeds \(72\,\text{km h}^{-1}\) and \(36\,\text{km h}^{-1}\) respectively. The train-A whistles at \(640\,\text{Hz}\) frequency. Before the trains meet, frequency of sound heard by a passenger in Train-B is \((v=340\,\text{m s}^{-1})\):

• A. \(500\,\text{Hz}\)
• B. \(600\,\text{Hz}\)
• C. \(700\,\text{Hz}\)
• D. \(800\,\text{Hz}\)

Hint:

In Doppler effect, when source and observer move towards each other, apparent frequency increases and is given by \[ f' = f\left(\frac{v+v_o}{v-v_s}\right) \]

Answer 1

The Correct Option is C

Step 1: Convert speeds to SI units.
Train A (source): \( v_s = 72\times\frac{5}{18} = 20\,\text{m s}^{-1} \). Train B (observer): \( v_o = 36\times\frac{5}{18} = 10\,\text{m s}^{-1} \). Both approach each other.

Step 2: Apply the Doppler formula.
\[ f' = f_0\cdot\frac{v+v_o}{v-v_s} = 640\times\frac{340+10}{340-20} = 640\times\frac{350}{320} = 700\,\text{Hz} \] \[ \boxed{700\,\text{Hz}} \]