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(D) Given,the point $A$ covers an angle $	heta = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$ in time $t = 0.1\, s$.Since the speed is uniform,the ang

Vedclass · Accepted Answer

$9.87$

Vedclass · Answer

(D) Given,the point $A$ covers an angle $	heta = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$ in time $t = 0.1\, s$.Since the speed is uniform,the angular velocity $\omega$ is given by $\omega = \frac{	heta}{t} = \frac{\pi/6}{0.1} = \frac{\pi}{0.6} = \frac{10\pi}{6} = \frac{5\pi}{3} 	ext{ rad/s}$.The radius of the circle $R = 0.36\, m$,which represents the amplitude $A_{amp}$ of the simple harmonic motion.The restoring force $F$ on a particle of mass $m$ executing simple harmonic motion is given by $F = -m\omega^2 x$,where $x$ is the displacement.When $P$ touches $M$,the displacement $x$ is equal to the amplitude $R = 0.36\, m$.The restoring force per unit mass is $\frac{F}{m} = \omega^2 R$.Substituting the values: $\frac{F}{m} = \left(\frac{5\pi}{3}ight)^2 	imes 0.36 = \frac{25\pi^2}{9} 	imes 0.36$.Using $\pi^2 \approx 9.87$,we get $\frac{F}{m} = \frac{25 	imes 9.87}{9} 	imes 0.36 = 25 	imes 9.87 	imes 0.04 = 25 	imes 0.3948 = 9.87\, N/kg$.

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