Question

A train moves towards a stationary observer with speed 72 m/s\(^{-1}\). The train blows its horn and its frequency heard by observer is \(f_1\). 
If the train speed is reduced to 36 m/s\(^{-1}\), the frequency heard by observer is \(f_2\). Then \( \frac{f_1}{f_2} \) is (given \(v = 340 { m/s}^{-1}\)):

• A. \(1.53\)
• B. \(1.43\)
• C. \(1.13\)
• D. \(1.23\)

Hint:

When calculating frequency changes due to the Doppler effect, always pay attention to the direction of the source relative to the observer to apply the correct formula.

Answer 1

The Correct Option is C

Step 1: Use the Doppler Effect formula for sound. 
The observed frequency \( f' \) when the source moves towards a stationary observer is given by: \[ f' = f \left(\frac{v}{v - v_s}\right), \] where \( f \) is the actual frequency of the sound, \( v \) is the speed of sound, and \( v_s \) is the speed of the source. 
Step 2: Calculate the ratio \( \frac{f_1}{f_2} \) for the different speeds of the train. \[ f_1 = f \left(\frac{340}{340 - 72}\right), \quad f_2 = f \left(\frac{340}{340 - 36}\right). \] Simplifying these: \[ f_1 = f \left(\frac{340}{268}\right) \approx f \times 1.269, \quad f_2 = f \left(\frac{340}{304}\right) \approx f \times 1.118. \] Thus, the ratio: \[ \frac{f_1}{f_2} = \frac{1.269}{1.118} \approx 1.135. \]