A wire of density $8 \times 10^3\,kg / m ^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If $Y =8 \times 10^{10}\,N / m ^2$, the fundamental frequency of vibration in the wire will be $......\,Hz$.
JEE MAIN 2023, Medium
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$f =\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}=\frac{1}{2 L } \sqrt{\frac{ YA \Delta L }{\rho A L}}$
$f =80\,Hz$
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