A granite rod of 60 cm length is clamped at | vedclass

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(D) In solids,the velocity of a longitudinal wave is given by $v = \sqrt{\frac{Y}{ho}}$.Substituting the given values: $v = \sqrt{\frac{9.27 	imes

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(D) In solids,the velocity of a longitudinal wave is given by $v = \sqrt{\frac{Y}{ho}}$.Substituting the given values: $v = \sqrt{\frac{9.27 	imes 10^{10}}{2.7 	imes 10^3}} = \sqrt{3.433 	imes 10^7} \approx 5859 \ m/s$.Since the rod is clamped at its middle,the middle point acts as a node $(N)$ and the ends act as antinodes $(A)$. For the fundamental mode,the length of the rod $L$ corresponds to half a wavelength,so $L = \frac{\lambda}{2}$,which implies $\lambda = 2L$.Given $L = 60 \ cm = 0.6 \ m$,we have $\lambda = 2 	imes 0.6 = 1.2 \ m$.The fundamental frequency is $f = \frac{v}{\lambda} = \frac{5859}{1.2} \approx 4882.5 \ Hz$.Converting to $kHz$,$f \approx 4.88 \ kHz$,which is approximately $5 \ kHz$.

$A$ 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 in $kHz$?

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