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An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now, it is closed at one end. If the frequency of the tuning fork is increased slowly from f1, -

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प्रश्न

An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now, it is closed at one end. If the frequency of the tuning fork is increased slowly from f1, then again a resonance is obtained with a frequency f2. If in this case the pipe vibrates nth harmonic, then ______.

विकल्प

  • n = 3, f1 = `3/4`f2

  • n = 3, f2 = `5/4`f1

  • n = 5, f2 = `5/4`f1

  • n = 5, f2 = `3/4`f1

MCQ
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उत्तर

An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now, it is closed at one end. If the frequency of the tuning fork is increased slowly from f1, then again a resonance is obtained with a frequency f2. If in this case the pipe vibrates nth harmonic, then `underlinebb("n" = 5, "f"_2 = 5/4"f"_1`.

Explanation:

Open pipe resonance frequency f= `(2"v")/(2"L")`

Closed pipe resonance frequency f= `"nv"/(4"L")`

f= `"n"/4`f1(where n is odd and f2 > f1)  

∴ n = 5

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