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A small, plane strip is suspended from a fixed support through a string of length $l$ as shown. A continuous beam of monochromatic light is incident horizontally on the strip and is completely absorbed. The energy falling on the strip per unit time is $P$ .
If the strip is deflected slightly from its equilibrium position in the plane of the figure, what will be the time period of the resulting oscillations?

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$2 π \sqrt{\frac{l}{\sqrt{g^{2} + {(\frac{P}{m c})}^{2}}}}$

$π \sqrt{\frac{l}{\sqrt{g^{2} + {(\frac{P}{m c})}^{2}}}}$

$2 π \sqrt{\frac{2 l}{\sqrt{g^{2} + {(\frac{P}{m c})}^{2}}}}$

$2 π \sqrt{\frac{l}{\sqrt{2 g^{2} + {(\frac{P}{m c})}^{2}}}}$

answer is A.

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Detailed Solution

Radiation force on the strip = $\frac{P}{c}$

Since angle of deflection of the string from vertical line is very small ,forces mg and $\frac{P}{c}$ are nearly mutually perpendicular.

Therefore tension in the string in equilibrium is given by

$T = {[(m g)^{2} + {(\frac{P}{c})}^{2}]}^{\frac{1}{2}}$

$\Rightarrow \frac{T}{m} = {[g^{2} + {(\frac{P}{m c})}^{2}]}^{\frac{1}{2}}$

This plays the role of effective $g$ . The time period of small oscillations is

$t = 2 π \sqrt{\frac{l}{T / m}} = 2 π \sqrt{\frac{l}{\sqrt{g^{2} + {(\frac{P}{m c})}^{2}}}}$

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