Oscillating disk under action of point source of light.

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Consider a thin ring of radius $\displaystyle x$ , at an angle $\displaystyle \theta$ from the point source.

$\tan\theta = \frac{x}{h} --- (1)$

Now, let us find the intensity of light at this distance $\displaystyle x$ .

$I = \frac{P}{4\pi (h\sec\theta)^2}$

The differential area of the thin ring is,

$dA = 2\pi xdx$

Also, the change in momentum of the light, after reflecting from the disk, is,

$\Delta p = 2p\cos\theta = 2\frac{h}{\lambda}\cos\theta --- (2)$

The number of photons striking the disk per unit time is,

$\eta = \frac{I\cdot dA\cos\theta}{\frac{hc}{\lambda}} --- (3)$

Using $\displaystyle (2)$ and $\displaystyle (3)$ , the differential force on the thin ring is,

$dF = \Delta p\times\eta = \frac{Px\cos^4\theta}{ch^2}dx$

Using $\displaystyle (1)$ ,

$dF = \frac{P\sin\theta\cos\theta}{c}d\theta$

Integrating with appropriate limits,

$F = \frac{PR^2}{2c(R^2+h^2)}$

For equilibrium,

$F = mg --- (4)$

Now, to find the time period, let us displace the disk by a small distance $\displaystyle dx$

$F_{net} = F - mg$

$\frac{dF_{net}}{dx} = \frac{dF}{dx} = \frac{-PR^2h}{c(R^2+h^2)^2}$

Thus,

$T = \frac{2\pi}{\sqrt{\frac{PR^2h}{mc(R^2+h^2)^2}}}$

Substituting for $\displaystyle m$ using $\displaystyle (4)$ , we get,

$T = 2\pi\sqrt{\frac{R^2+h^2}{2gh}}$

Substituting the values provided,

$\boxed{T = 2.244\text{sec}}$