Question

A train moving at 20 m/s approaches a stationary observer. The frequency of the whistle emitted by the train is 640 Hz. If the velocity of sound is 340 m/s, the apparent frequency heard by the observer is:

• A. 680 Hz
• B. 600 Hz
• C. 720 Hz
• D. 640 Hz

Hint:

Identify which object is moving, the source or the observer, and in which direction, since this decides whether you add or subtract the object speed in the Doppler formula. Keep the speed of sound and every object speed in the same units before substituting.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem uses the Doppler Effect. Since the source is moving toward a stationary observer, the frequency heard will be higher than the actual frequency.
Key Formula or Approach:
Doppler Effect formula: \( f_{app} = f \left( \frac{v \pm v_o}{v \mp v_s} \right) \).
For a stationary observer (\( v_o = 0 \)) and source approaching (\( v - v_s \) in denominator):
\[ f' = f \left( \frac{v}{v - v_s} \right) \] Step 2: Detailed Explanation:
Given:
Actual frequency \( f = 640 \text{ Hz} \).
Velocity of sound \( v = 340 \text{ m/s} \).
Velocity of source (train) \( v_s = 20 \text{ m/s} \).
Apply the formula:
\[ f' = 640 \left( \frac{340}{340 - 20} \right) \] \[ f' = 640 \left( \frac{340}{320} \right) \] \[ f' = 640 \times 1.0625 = 2 \times 340 = 680 \text{ Hz} \] (Alternatively, \( 640/320 = 2 \), then \( 2 \times 340 = 680 \)).
Step 3: Final Answer:
The apparent frequency is 680 Hz.
This matches Option (A).