Question

Two particles $P$ and $Q$ are executing SHM across same straight line whose equations are given as $y _{ P }= A \sin \left(\omega t +\phi_{1}\right)$ and $y _{ Q }= A \cos \left(\omega t +\phi_{2}\right)$. An observer, at $t =0$, observers the particle $P$ at a distance $A / \sqrt{2}$ moving to the right from mean position $O$ while $Q$ at $\frac{\sqrt{3}}{2}$ A moving to the left from mean position O as shown. Then, $\left(\phi_{2}-\phi_{1}\right)$ is equal to

• A. $\pi / 12$
• B. $7 \pi / 12$
• C. $5 \pi / 12$
• D. None of the above

Answer 1

Answer: (B)

$y _{ P }= A \sin \left(\omega t +\phi_{1}\right)$
at $t =0$
$\frac{ A }{\sqrt{2}}= A \sin \phi_{1} $
$\Rightarrow \sin \phi_{1}=\frac{1}{\sqrt{2}}$
$ \Rightarrow \phi_{1}=\pi / 4,3 \pi / 4$
$\& v _{ P }= A \omega \cos \phi_{1}$
since $v _{ P }=(+)$ ive so, $\phi_{1}=\pi / 4$
$y _{ Q }= A \cos \left(\omega t +\phi_{2}\right)$
at $t =0$
$-\frac{\sqrt{3} A }{2}= A \cos \phi_{2} $
$\Rightarrow \cos \phi_{2}=-\frac{\sqrt{3}}{2}$
$ \Rightarrow \phi_{2}=\pi-\pi / 6=5 \pi / 6$
$\pi+\pi / 6=7 \pi / 6$
$\& v _{ Q }=- A\omega \sin \phi_{2}$,
since $v _{ Q }=(-)$ ive so, $\phi_{2}=5 \pi / 6$
$\therefore \Delta \phi=5 \pi / 6-\pi / 4$
$=\frac{10 \pi-3 \pi}{12}=7 \pi / 12$