Question

Two particles $P$ and $Q$ are executing SHM across same straight line whose equations are given as $y_P=A\sin \left(\omega t+{\varphi }_1\right)$ and $y_Q=A\cos \left(\omega t+{\varphi }_2\right)$. 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, $\left({\varphi }_2-{\varphi }_1\right)$ is equal to

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

Answer 1

Answer: (B)

$y_P=A\sin \left(\omega t+{\varphi }_1\right)$
at $t=0$
$\frac{A}{\sqrt{2}}=A\sin {\varphi }_1$
$\Rightarrow \sin {\varphi }_1=\frac{1}{\sqrt{2}}$
$\Rightarrow {\varphi }_1=\pi /4,3\pi /4$
$\&v_P=A\omega \cos {\varphi }_1$
since $v_P=(+)$ ive so, ${\varphi }_1=\pi /4$
$y_Q=A\cos \left(\omega t+{\varphi }_2\right)$
at $t=0$
$-\frac{\sqrt{3}A}{2}=A\cos {\varphi }_2$
$\Rightarrow \cos {\varphi }_2=-\frac{\sqrt{3}}{2}$
$\Rightarrow {\varphi }_2=\pi -\pi /6=5\pi /6$
$\pi +\pi /6=7\pi /6$
$\&v_Q=-A\omega \sin {\varphi }_2$,
since $v_Q=(-)$ ive so, ${\varphi }_2=5\pi /6$
$\therefore \Delta \varphi =5\pi /6-\pi /4$
$=\frac{10\pi -3\pi }{12}=7\pi /12$