Question

A particle executes $\text{SHM}$ of amplitude $A$ and time period $T$. Find the distance travelled by the particle in the duration its phase changes by $\frac{\pi }{12}$ to $\frac{5\pi }{12}$.

• A. $\frac{A}{\sqrt{2}}$
• B. $\sqrt{\frac{3}{2}}A$
• C. $\frac{2}{\sqrt{3}}A$
• D. $\sqrt{\frac{2}{3}}A$

Answer 1

The correct option is A $\frac{A}{\sqrt{2}}$

Initial phase $\left({\delta }_1\right)=\frac{\pi }{12}={15}^{\circ }$
Final phase $\left({\delta }_2\right)=\frac{5\pi }{12}={75}^{\circ }$
As both ${\delta }_1$ and ${\delta }_2$ are acute.
Therefore, Distance travelled by the particle = Displacement by the particle
$\Rightarrow x=A\sin {\delta }_2-A\sin {\delta }_1=A\sin \frac{5\pi }{12}-A\sin \frac{\pi }{12}$
$\because \sin C-\sin D=2\cos \left(\frac{C+D}{2}\right).\sin \left(\frac{C-D}{2}\right)$
$\Rightarrow x=2A\cos \frac{\pi }{4}\sin \frac{\pi }{6}=\frac{A}{\sqrt{2}}$