Question

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

• A. $100\text{N}$
• B. $50\text{N}$
• C. $9.87\text{N}$
• D. $0.49\text{N}$

Answer 1

The correct option is C $9.87\text{N}$


The point $A$ covers ${30}^{\circ }$ in $0.1\text{s}$.
From unitary method it means,
$\frac{\pi }{6}\to 0.1\text{s}$
$1\to \frac{0.1}{\frac{\pi }{6}}\text{s}$
$2\pi \to \frac{0.1}{\frac{\pi }{6}}\times 2\pi \text{s}$
$T=1.2\text{s}$

We know that $\omega =\frac{2\pi }{T}=\frac{2\pi }{1.2}$

Restoring force $\left(F\right)=m{\omega }^{2}A$

Then, restoring force per unit mass $\left(\frac{F}{m}\right)={\omega }^{2}A={\left(\frac{2\pi }{1.2}\right)}^2\times 0.36\approx 9.87\text{N}$