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JEE Mains · Physics · STD 11 - 14. waves and sound

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 .... \(kHz\) ?

  1. A \(2.5\)
  2. B \(10\)
  3. C \(7.5\)
  4. D \(5\)
Verified Solution

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…
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