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:

Source moving towards observer $\Rightarrow$ use minus in denominator ($v - v_s$) $\Rightarrow$ frequency increases. Source moving away $\Rightarrow$ plus in denominator $\Rightarrow$ frequency decreases.

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).