Question:
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :
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For source speed \(u\ll v\): \(f_{app}\approx f_0\left(1\pm\dfrac uv\right)\), so the beat frequency is \(\dfrac{2f_0u}{v}\).
Updated On: Oct 31, 2026
- 1 m/s
- \( {1 \over 8} \) m/s
- \( {1 \over 4} \) m/s
- \( {1 \over 2} \) m/s
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The Correct Option is C
Solution and Explanation
Step 1: Approaching fork: \(f_1=\dfrac{f_0v}{v-u}\approx f_0\left(1+\dfrac uv\right)\). Receding fork: \(f_2=\dfrac{f_0v}{v+u}\approx f_0\left(1-\dfrac uv\right)\).
Step 2: Beat frequency \(=f_1-f_2=\dfrac{2f_0u}{v}=2\).
Step 3: \(u=\dfrac{2v}{2f_0}=\dfrac{350}{1400}=0.25\) m/s \(=\dfrac14\) m/s.
Answer: (C) \(\dfrac14\) m/s
Step 2: Beat frequency \(=f_1-f_2=\dfrac{2f_0u}{v}=2\).
Step 3: \(u=\dfrac{2v}{2f_0}=\dfrac{350}{1400}=0.25\) m/s \(=\dfrac14\) m/s.
Answer: (C) \(\dfrac14\) m/s
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