An object of mass ‘m’ is suspended at the end of a massless wire of length L and area of cross-section, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is :

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$f = \frac{1}{2 π} \sqrt{\frac{m A}{Y L}}$

$f = \frac{1}{2 π} \sqrt{\frac{Y L}{m A}}$

$f = \frac{1}{2 π} \sqrt{\frac{Y A}{m L}}$

$f = \frac{1}{2 π} \sqrt{\frac{m L}{Y A}}$

answer is C.

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Detailed Solution

$\begin{array}{l} Wire acts like a spring . \\ F = kx = (\frac{YA}{L}) △ L \\ \Rightarrow Spring constant = k = \frac{YA}{L} \\ ∴ Frequency of SHM = f = \frac{1}{2 π} \sqrt{\frac{k}{m}} = \frac{1}{2 π} \sqrt{\frac{YA}{mL}} \end{array}$

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