An object of mass m is suspended at the end of a massless wire of length L and area of crosssection 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 :

f = 1 \over 2\pi \sqrt YA \over mL

f = 1 \over 2\pi \sqrt mL \over YA

f = 1 \over 2\pi \sqrt YL \over mA

f = 1 \over 2\pi \sqrt mA \over YL

Properties of Matter PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q43 with Solution

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Correct answer: A

When a wire is stretched by a force $F$ , its extension $x$ is given by Young's modulus relation:

Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L}

So,

Y = \frac{FL}{Ax}

Rearranging,

F = \frac{YA}{L}x

This is of the form $F = kx$ , hence the effective spring constant of the wire is

k = \frac{YA}{L}

If the mass $m$ is pulled down slightly and released, it performs small oscillations about the equilibrium position.

For a mass-spring system, the angular frequency is

\omega = \sqrt{\frac{k}{m}}

Substituting $k = \dfrac{YA}{L}$ ,

\omega = \sqrt{\frac{YA}{mL}}

Now frequency,

f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}

The obtained result is

f = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}

This matches Option A.

\boxed{f = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}}