A uniform plate of mass M rests horizontally and symmetrically on...

Let the center of mass (CM) of the plate be displaced by a distance $x$ towards the right from its equilibrium position.

The distance of the CM from the left wheel becomes $\dfrac{L}{2} + x$ and from the right wheel becomes $\dfrac{L}{2} - x$.

Let $N_1$ and $N_2$ be the normal reactions from the left and right wheels, respectively. For vertical equilibrium:

$N_1 + N_2 = Mg$

For rotational equilibrium about the CM, the net torque is zero:

$N_1$\left( \dfrac{L}{2} + x \right) = N_2 $\left( \dfrac{L}{2} - x \right)$

Solving these two equations, we get:

$N_1 = \dfrac{Mg}{L} \left( \dfrac{L}{2} - x \right)$ and $N_2 = \dfrac{Mg}{L} \left( \dfrac{L}{2} + x \right)$

The wheels are rotating inwards, so the kinetic friction forces act towards the center. The left wheel exerts a force $f_1 = \mu N_1$ towards the right, and the right wheel exerts a force $f_2 = \mu N_2$ towards the left.

The net restoring force on the plate is:

$F = f_1 - f_2 = \mu N_1 - \mu N_2 = \mu (N_1 - N_2)$

Substituting the values of $N_1$ and $N_2$:

$F = \mu \dfrac{Mg}{L} \left[ \left( \dfrac{L}{2} - x \right) - \left( \dfrac{L}{2} + x \right) \right] = -\dfrac{2 \mu Mg}{L} x$

The acceleration of the plate is:

$a = \dfrac{F}{M} = -\left( \dfrac{2 \mu g}{L} \right) x$

This represents simple harmonic motion with $\omega^2 = \dfrac{2 \mu g}{L}$.

The time period of oscillation is:

$T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{L}{2 \mu g}}$