A rod of mass ' M ' and length ' 2 L ' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of ' $m$ ' are attached at distance ' $\mathrm{L} / 2$ ' from its centre on both sides, it reduces the oscillation frequency by $20 \%$. The value of ratio $\mathrm{m} / \mathrm{M}$ is close to:

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Elasticity & SHM

A rod of mass ' M ' and length ' 2 L ' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of '

m

' are attached at distance '

L /2

' from its centre on both sides, it reduces the oscillation frequency by

20%

. The value of ratio

m / M

is close to:

Answer: (C)

Treat this as angular SHM, where the frequency is

f = \frac{1}{2 π} \frac{C}{I}

with

C

the torsion constant of the wire and

I

the moment of inertia about the suspension axis. The wire is unchanged, so

f \propto \frac{1}{I}

. For the bare rod of length

2 L

about its centre,

I_{1} = \frac{M ( 2 L ) ^{2}}{12} = \frac{M L ^{2}}{3}

. Hanging a mass

m

\frac{L}{2}

on each side adds

2 m (\frac{L}{2})^{2} = \frac{m L ^{2}}{2}

, so

I_{2} = \frac{M L ^{2}}{3} + \frac{m L ^{2}}{2}

. A

20%

drop in frequency means

f_{2} = 0.8 f_{1}

, hence

\frac{I _{2}}{I _{1}} = (\frac{f _{1}}{f _{2}})^{2} = \frac{25}{16}

. Writing this out,

1 + \frac{3 m}{2 M} = \frac{25}{16}

, so

\frac{3 m}{2 M} = \frac{9}{16}

and

\frac{m}{M} = \frac{3}{8} = 0.375

, which rounds to about

0.37

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option