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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=υ4×40×10−2 and f2=υ4 L2=υ4×41×10−2 we can see f1>f2 ∴f1−f2=10 ...(i) and f1f2=4×41×10−24×40×10−2=4140 ...(ii) putting the value of f2 in eq (i) So, f1−4041f1=10 ⇒f141=10⇒f1=410 Hz ∴ f2=410−10=400 Hz

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