QJEEAdv08Ph9529: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ph9529 Solution: The centre of mass of the disk undergoes simple harmonic motion …

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[ $\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$- 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$\Rightarrow {3 \over 2}M{a_c} = 2kx$

$\Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $= {{ - 4kx} \over 3}$ directed towards the equilibrium.

$\therefore$ $|{F_{net}}| = {{4k} \over 3}x$

For S.H.M $|{F_{net}}| = M{\omega ^2}x$

$\therefore$ $M{\omega ^2} = {{4k} \over 3}$

$\Rightarrow \omega = \sqrt {{{4k} \over {3M}}}$ .... (iii)