As shown in the figure, the radius of the pulley is R, its moment of inertia relative to the rotation axis is I, the mass of the body is m, and the spring constant is K. The mass of the thread and the spring is negligible, the thread does not slide over the pulley, and there is no friction in the axis of the pulley, then

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The frequency of oscillation = $\frac{1}{2 π} \sqrt{\frac{K}{I / R^{2} + m}}$

Angular acceleration = $\frac{K R^{2} θ}{I + m R^{2}}$

The time period of oscillation = $2 π \sqrt{\frac{I + m R^{2}}{K R^{2}}}$

The angular frequency = $\sqrt{\frac{m + I / R^{2}}{K}}$

answer is A, B, C.

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Detailed Solution

When the spring is displaced by a small amount, force on the spring is k x. So, the net torque on the pulley = $K x . R$
But $x = R θ$
or net torque = $K R^{2} θ = - I \frac{d^{2} θ}{d t^{2}} = (1 + m R^{2}) \frac{d^{2} θ}{d t^{2}} ∴ \frac{d^{2} θ}{d t^{2}} = \frac{K R^{2} θ}{(1 + m R^{2})}$
Angular acceleration is proportional to angular displacement. So, it executes angular SHM.
$ω = \sqrt{\frac{K}{1 / R^{2} + m}}$
$∴$ Period $T = 2 π \sqrt{\frac{1 + m R^{2}}{K R^{2}}}$
Frequency = $\frac{1}{T} = \frac{1}{2 π} \sqrt{\frac{K R^{2}}{1 + m R^{2}}} = \frac{1}{2 π} \sqrt{\frac{R}{1 / R^{2} + m}}$