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 $2.7\times {10}^3$ 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?

• A. $10$ kHz
• B. $7.5$ kHz
• C. $5$ kHz
• D. $2.5$ kHz

Answer 1

Answer: (C)

$5$ kHz

Given : $Y=9.27\times {10}^{10}Pa$ $\rho =2.7\times {10}^{3}kg/m^3$
Length of rod $L=60cm=0.6m$
Velocity of sound in granite $v=\sqrt{\frac{Y}{\rho }}=\sqrt{\frac{9.27\times {10}^{10}}{2.7\times {10}^3}}=5.9\times {10}^{3}m/s$
Fundamental frequency $\nu =\frac{v}{2L}=\frac{5.9\times {10}^3}{2\times 0.6}=5kHz$