Question:
A train moves towards a stationary observer with speed \(20\) m/s. The train sounds a whistle and its frequency heard by the observer is \(f_{1}\). When the train moves away from the platform at a speed of \(20\) m/s, the frequency heard is \(f_{2}\). If the speed of sound is \(340\) m/s, then \(\frac{f_{2}}{f_{1}}\) is
A train moves towards a stationary observer with speed \(20\) m/s. The train sounds a whistle and its frequency heard by the observer is \(f_{1}\). When the train moves away from the platform at a speed of \(20\) m/s, the frequency heard is \(f_{2}\). If the speed of sound is \(340\) m/s, then \(\frac{f_{2}}{f_{1}}\) is
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Use the Doppler formula for a moving source and a stationary observer, with minus for approach and plus for recession.
Updated On: Nov 12, 2026
- \(\frac{7}{8}\)
- \(\frac{8}{7}\)
- \(\frac{9}{8}\)
- \(\frac{8}{9}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For a moving source and a stationary observer, \(f'=f\dfrac{v}{v\mp v_s}\), with minus when the source approaches.
Step 2: Approaching
\(f_1=f\dfrac{340}{340-20}=f\dfrac{340}{320}\).
Step 3: Receding
\(f_2=f\dfrac{340}{340+20}=f\dfrac{340}{360}\).
Step 4: Result
\(\dfrac{f_2}{f_1}=\dfrac{320}{360}=\dfrac{8}{9}\), which is less than 1 because the pitch drops as the train goes away.
Final Answer:
The ratio f2 to f1 is 8/9. \[ \boxed{\text{(D) }\dfrac{8}{9}} \]
For a moving source and a stationary observer, \(f'=f\dfrac{v}{v\mp v_s}\), with minus when the source approaches.
Step 2: Approaching
\(f_1=f\dfrac{340}{340-20}=f\dfrac{340}{320}\).
Step 3: Receding
\(f_2=f\dfrac{340}{340+20}=f\dfrac{340}{360}\).
Step 4: Result
\(\dfrac{f_2}{f_1}=\dfrac{320}{360}=\dfrac{8}{9}\), which is less than 1 because the pitch drops as the train goes away.
Final Answer:
The ratio f2 to f1 is 8/9. \[ \boxed{\text{(D) }\dfrac{8}{9}} \]
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