Question:
A train is moving towards a stationary observer with speed \(34\) m/s. A train sounds a whistle of frequency \(450\) Hz. If the speed of sound is \(340\) m/s, the frequency heard by the observer in Hz is
A train is moving towards a stationary observer with speed \(34\) m/s. A train sounds a whistle of frequency \(450\) Hz. If the speed of sound is \(340\) m/s, the frequency heard by the observer in Hz is
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Source moving toward a stationary observer: f = f0 v/(v - vs).
Updated On: Oct 1, 2026
- \(440\)
- \(480\)
- \(500\)
- \(540\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
When a source moves toward a stationary observer, the observed frequency increases because the wavefronts are compressed.
Step 2: Formula:
\[ f' = f\cdot\frac{v}{v - v_s} \]
Step 3: Compute:
\[ f' = 450\times\frac{340}{340 - 34} = 450\times\frac{340}{306} = 450\times\frac{10}{9} = 500 \text{ Hz} \]
Step 4: Why the other options are wrong.
440 Hz is lower than the source frequency, which would be heard if the train were moving away. 480 Hz and 540 Hz do not match \(\frac{340}{306}\), and 540 Hz would be \(450\times1.2\).
Final Answer:
The frequency heard is \(500\) Hz, option (C). \[ \boxed{500\text{ Hz}} \]
When a source moves toward a stationary observer, the observed frequency increases because the wavefronts are compressed.
Step 2: Formula:
\[ f' = f\cdot\frac{v}{v - v_s} \]
Step 3: Compute:
\[ f' = 450\times\frac{340}{340 - 34} = 450\times\frac{340}{306} = 450\times\frac{10}{9} = 500 \text{ Hz} \]
Step 4: Why the other options are wrong.
440 Hz is lower than the source frequency, which would be heard if the train were moving away. 480 Hz and 540 Hz do not match \(\frac{340}{306}\), and 540 Hz would be \(450\times1.2\).
Final Answer:
The frequency heard is \(500\) Hz, option (C). \[ \boxed{500\text{ Hz}} \]
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