JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Elasticity & SHM

An object of mass

m

is suspended at the end of a massless wire of length

L

and area of cross - selection, 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:

(A) $f = \frac{1}{2 π} \frac{mL}{YA}$
(B) $f = \frac{1}{2 π} \frac{YA}{mL}$ ✓ Correct
(C) $f = \frac{1}{2 π} \frac{mA}{YL}$
(D) $f = \frac{1}{2 π} \frac{YL}{mA}$

Answer: (B)

Solution

A loaded wire behaves like a spring: stretching it by

x

develops a restoring force

\frac{Y A}{L} x

, so its effective stiffness is

k = \frac{Y A}{L}

. A mass

m

on a spring of stiffness

k

oscillates at frequency

f = \frac{1}{2 π} \frac{k}{m}

. Substituting

k = \frac{Y A}{L}

gives

f = \frac{1}{2 π} \frac{Y A}{m L}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option

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