The point A moves with a uniform speed along the circumference of a circle of radius $0.36 \mathrm{~m}$ and cover $30^{\circ}$ in $0.1 s$. The perpendicular projection 'P' form ' $A^{\prime}$ on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be:
Correct Option: , 3
The point a covers $30^{\circ}$ in $0.1$ sec.
Means $\frac{\pi}{6} \longrightarrow 0.1 \mathrm{sec}$
$1 \longrightarrow \frac{0.1}{\frac{\pi}{6}}$
$2 \pi=\longrightarrow \frac{0.1 \times 6}{\pi} \times 2 \pi$
$\mathrm{T}=1.2 \mathrm{sec}$
We know that $\omega=\frac{2 \pi}{\mathrm{T}}$
$\omega=\frac{2 \pi}{1.2}$
Restoration force $(F)=m \omega^{2} A$
Then Restoration force per unit mass $\left(\frac{F}{\mathrm{w}}\right)=\omega^{2} \mathrm{~A}$
$\left(\frac{F}{m}\right)=\left(\frac{2 \pi}{1.2}\right)^{2} \times 0.36$
$\cong 9.87 \mathrm{~N}$