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Question

In a resonance tube experiment, tuning fork A is in resonance with air column of length 40 cm and tuning fork B is in resonance with air column of length 41 cm.
Two tuning forks A and B sounded together produce 10 beats per second. Calculate the frequencies of forks.
(Assume fundamental mode for air column)

A
400 Hz,390 Hz
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B
430 Hz,410 Hz
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C
410 Hz,400 Hz
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D
410 Hz,420 Hz
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Solution

The correct option is C 410 Hz,400 Hz
Let the frequency of the first fork be f1 and that of second be f2.
For tube closed at one end,
f1=υ4 L1=υ4×40×102
and
f2=υ4 L2=υ4×41×102
we can see
f1>f2
f1f2=10 ...(i)
and
f1f2=4×41×1024×40×102=4140 ...(ii)
putting the value of f2 in eq (i)
So,
f14041f1=10
f141=10f1=410 Hz
f2=41010=400 Hz

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