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 \(340\, ms^{-1}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(200\)
Step-by-step Solution
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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