Question

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Q. 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 ?
(1) 7.5 kHz
(2) 5 kHz
(3) 2.5 kHz
(4) 10 kHz

Answer 1

$$\begin{gathered} Dearstudent, \\ velocityofwave=\sqrt{\frac{Y}{\rho }}=\sqrt{\frac{9.27\times {10}^{10}}{2.7\times {10}^3}}=5.85\times {10}^{3}m/s \\ asrodisclampedatmiddlehence, \\ \frac{\lambda }{2}=L \\ \lambda =2L \\ L=60cm=0.6m \\ \lambda =1.2m \\ v=f\lambda \\ f=\frac{5.85\times {10}^3}{1.2}=4.88KHz\approx 5KHz \\ Regards \end{gathered}$$