The string and the spring depicted in the figure are light....

Let the mass $m$ be displaced downwards by a small distance $x$ from its equilibrium position. The velocity of the mass is $v = \dot{x}$.

Since the string is inextensible and does not slip, the pulley rotates by an angle $\theta = x/r$, and its angular velocity is $\omega = v/r$. The spring is stretched by an additional amount $x$.

The total mechanical energy of the system is:

$E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}k(x_0 + x)^2 - mgx$

where $x_0$ is the initial extension of the spring in equilibrium, satisfying $kx_0 = mg$.

Substituting $\omega = v/r$ and expanding the potential energy term:

$E = \dfrac{1}{2} \left(m + \dfrac{I}{r^2}\right) v^2 + \dfrac{1}{2}k x_0^2 + k x_0 x + \dfrac{1}{2}k x^2 - mgx$

Since $kx_0 = mg$, the terms $k x_0 x$ and $-mgx$ cancel out:

$E = \dfrac{1}{2} \left(m + \dfrac{I}{r^2}\right) v^2 + \dfrac{1}{2}k x^2 + \text{constant}$

Differentiating the total energy with respect to time ($\dfrac{dE}{dt} = 0$) gives:

\left(m + \dfrac{I}{r^2}\right) v \dfrac{dv}{dt} + k x \dfrac{dx}{dt} = 0$<br><br>Using$v = \dfrac{dx}{dt}$and$a = \dfrac{dv}{dt}$, we get:<br><br>\left(m + \dfrac{I}{r^2}\right) a + k x = 0 \Rightarrow a = -\dfrac{k}{m + I/r^2} x$

Comparing this with the standard equation of SHM $a = -\omega^2 x$, the angular frequency is:

$\omega = \sqrt{\dfrac{k}{m + I/r^2}}$

The time period of oscillation is:

$T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{m + I/r^2}{k}}$