Question:
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7×10^3 kg/m^3$ and its Young's modulus is $9.27×10^{10} Pa$. What will be the fundamental frequency of the longitudinal vibrations?
A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7×10^3 kg/m^3$ and its Young's modulus is $9.27×10^{10} Pa$. What will be the fundamental frequency of the longitudinal vibrations?
Updated On: Jun 26, 2024
2.5KHz
5 KHz
7.5 KHz
10 KHz
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The Correct Option is B
Solution and Explanation
The correct answer is option (B): 5 KHz
Given:-
Y=$9.27×10^{10 }Pa$
ρ=$2.7×10^3 kg/m^3$
Length of rod L=60 cm=0.6 m
Fundamental frequency = $\frac1{2L} \sqrt{\frac{Y}{ρ}}$
= $\frac1{2(0.6)} \sqrt{\frac{9.27×10^{10 }}{2.7×10^3 }}$
==$4.9×10^3 Hz = 5kHz $
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