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JEE Mains · Physics · STD 11 - Waves and sound
  • JEE Mains 2018
  • Hard
  • MCQ

A granite rod of \(60\ cm\) length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is \(2.7 \times 10^3 \) \(kg/m^3\) and its Young's modulus is \(9.27 \times 10^{10}\) \(Pa\) What will be the fundamental frequency of the longitudinal vibrations .... \(kHz\) ?

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Answer & Solution

Correct Answer

(D) \(5\)

Step-by-step Solution

Detailed explanation

In solids, Velocity of wave \(V=\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}\)

\(\mathrm{v}=5.85 \times 10^{3} \mathrm{m} / \mathrm{sec}\)

since rod is clamped at middle fundamental wave shape is as follow

\(\frac{\lambda}{2}=\mathrm{L} \Rightarrow \lambda=2 \mathrm{L}\)

\(\lambda=1.2 \mathrm{m}(\because \mathrm{L}=60 \mathrm{cm}=0.6 \mathrm{m}(\mathrm{given})\)

Using \(v=f \lambda\)

\(\Rightarrow \quad f=\frac{v}{\lambda}=\frac{5.85 \times 10^{3}}{1.2}\)

\(=4.88 \times 10^{3} \mathrm{Hz}=5 \mathrm{KHz}\)

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