JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Elasticity & SHM

A uniform thin cylindrical disk of mass

M

and radius

R

is attached to two identical massless springs of spring constant

k

which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance

d

from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is

L

. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance

L

from the wall. The disk rolls without slipping with velocity

V_{0} = V_{0} \hat{i}

. The coefficient of friction is

μ

.

The centre of mass of the disk undergoes simple harmonic motion with angular frequency

ω

equal to

Answer: (D)

Solution

From the previous question, we found that the net force is

F = - \frac{4 k x}{3}

.

For SHM,

F = - M ω^{2} x

.

Comparing:

M ω^{2} x = \frac{4 k x}{3}

ω^{2} = \frac{4 k}{3 M}

ω = \frac{4 k}{3 M}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option