# An object of mass 'm' is suspended at the end of a mass less wire of length L and area of cross-section A. Young's modules of the maternal of the wire is Y. if the mass is pulled down slightly its freq. of oscillation along the vertical direction is : Option: 1 Option: 2 Option: 3 Option: 4

At equilibrium

$T=mg$ ......(i)

$Y=\frac{T / A}{l / L}$

$\Rightarrow T=\frac{Y A l}{L} \ldots(ii)$

From
(i) and (ii)

$m g=\frac{Y A l}{L} \ldots (iii)$

Now for a small displacement x

\begin{aligned} &\text { Restoring force } &\left.=-\left[T^{\prime}-m g\right]=-\left[\frac{Y A(l+x)}{L}-\frac{Y A l}{L}\right] \right] &=\frac{-Y A x}{L} \end{aligned}Comparing with  $F=-m \omega^{2} x$

$\begin{array}{l} m \omega^{2}=\frac{Y A}{L}\\ \Rightarrow \omega=\sqrt{\frac{Y A}{m L}} \Rightarrow \frac{2 \pi}{T}=\sqrt{\frac{Y A}{m L}} \\ \text { Frequency } =f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}} \end{array}$

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