The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers $30^{0}$ in 0.1s. The perpendicular projection ‘P’ from ‘A’ on the diameter MN represents the simple harmonic motion of ‘P’. The restoration force per unit mass when P touches M will be:

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0.49 N

50 N

9.87 N

100 N

answer is B.

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Detailed Solution

$Amplitude = radius of circle=0.36m$

$Angular frequency = Angular speed= \frac{θ}{t} = \frac{\frac{π}{6}}{0.1} = \frac{π}{0.6} s$

$When P touches M,$

$\begin{array}{l} F_{max} = m a_{max} \\ \Rightarrow F_{max} = m ω^{2} A \\ \Rightarrow \frac{F_{max}}{m} = {(\frac{π}{0.6})}^{2} \times 0.36 = 9.87 N \end{array}$

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