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 .... \(kHz\) ?
- A \(2.5\)
- B \(10\)
- C \(7.5\)
- D \(5\)
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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