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 .... $kHz$ ?
JEE MAIN 2018, Diffcult
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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 $v=f \lambda$

$\Rightarrow \quad f=\frac{v}{\lambda}=\frac{5.85 \times 10^{3}}{1.2}$

$=4.88 \times 10^{3} \mathrm{Hz}=5 \mathrm{KHz}$

art

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