Question

Assertion :Time taken by a particle in SHM to move from x=A to $x=\frac{\sqrt{3}A}{2}$ is same as the time taken by the particle to move from $x=\frac{\sqrt{3}A}{2}$ to $x=\frac{A}{2}$. Reason: Corresponding angles rotated in the reference circle are same in the given time intervals.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

At the instant $t=0$ , the particle is at $x=A$ with phase equal to zero. Let the time taken by the particle to go from A to B be $t_1$ and from B to C be $t_2$.

The displacement equation is given by $x=Acoswt$

When it is at B i.e $x=\frac{\sqrt{3}A}{2}$, $\frac{\sqrt{3}A}{2}=Acosw{t}_1⟹w{t}_1=\frac{\pi }{6}={30}^o={\theta }_1$

$\therefore t_1=\frac{\pi }{6w}=\frac{T}{12}$ $(asw=\frac{2\pi }{T})$

When it is at C i.e $x=\frac{A}{2}$, $\frac{A}{2}=Acosw{t}^{\prime }⟹w{t}^{\prime }=\frac{\pi }{3}={60}^o$

Thus the angle rotated from going B to C ${\theta }_2=w{t}_2={60}^o-{30}^o={30}^o$

Thus time taken to move from B to C, $t_2=\frac{\pi }{6w}=\frac{T}{12}$