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 .... \(kHz\) ?
Pick an option to check your answer.
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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