Question 45 of 194

The point $A$ moves with a uniform speed along the circumference of a circle of radius $0.36m$ and covers $30^{\circ}$ 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 :

A$0.49N$B$9.87N$C$50N$D$100N$

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Answer: (B) $9.87N$

The point $A$ covers $30^{\circ}$ in $0.1s$.
Then, we have $\dfrac{\pi }{6}rad⟶0.1s$.
Or $1rad⟶\dfrac{0.1}{\dfrac{\pi }{6}}s$
Or $2\pi in\rightarrow \dfrac{0.1\times 6}{\pi }\times 2\pi$
Thus, time taken is $T=1.2\sec .$
We know that $\omega =\dfrac{2\pi }{T}$
Then, $\omega =\dfrac{2\pi }{1.2}$
Restoration force $(F)={m\omega }^{2}A.$
Then Restoration force per unit mass $\left(\dfrac{F}{m}\right)={\omega }^{2}A$
$\left(\dfrac{F}{m}\right)={\left(\dfrac{2\pi }{1.2}\right)}^{2}\times 0.36$
$\cong 9.87N$