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JEE Mains · Physics · STD 11 - Waves and sound
  • JEE Mains 2018
  • Easy
  • Numerical

A tuning fork vibrates with frequency \(256\, Hz\) and gives one beat per second with the third normal mode of vibration of an open pipe . What is the length of the pipe? ... \(cm\) (Speed of sound of air is \(340\, ms^{-1}\))

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Answer & Solution

Correct Answer

(D) \(200\)

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Detailed explanation

According to question, tuning fork gives \(1\, beat/second\) with \((N)\, 3^{rd}\) normal mode. Therefore, organ pipe will have frequency \(\left( {256 \pm 1} \right)\,Hz\). In open organ pipe, frequency

\(\mathrm{n}=\frac{\mathrm{NV}}{2 \ell}\)

or, \(255=\frac{3 \times 340}{2 \times \ell} \Rightarrow \ell=2 \mathrm{m}=200 \mathrm{cm}\)

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