Question

Two particles move parallel to $x-$ axis about the origin with same amplitude ${}^{\prime }{a}^{\prime }$ and frequency $\omega$. At a certain instant they are found at a distance $a/3$ from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two?

• A. ${\cos }^{-1}\frac{7}{9}$
• B. ${\cos }^{-1}\frac{5}{9}$
• C. ${\cos }^{-1}\frac{4}{9}$
• D. ${\cos }^{-1}\frac{1}{9}$

Answer 1

Answer: (A)

${\cos }^{-1}\frac{7}{9}$

Let $x_1=a\sin \omega t$ and $x_2=a\sin (\omega t+\delta )$ be two $SHMs$.
$\frac{a}{3}=a\sin \omega t$ and $-\frac{a}{3}=a\sin (\omega t+\delta )$
$\sin \omega t=1/3$ and $\sin (\omega t+\delta )=-1/3$
$\sin (\omega t+\delta )=\sin \omega t\cdot \cos \delta +\cos \omega t\cdot \sin \delta$

On puting the valus of $\sin \omega t\&\cos \omega t$

$\frac{1}{3}\cos \delta +\sqrt{1-\frac{1}{9}}\sin \delta =-\frac{1}{3}$
$9{\cos }^2\delta +2\cos \delta -7=0$
$\cos \delta =-1$ or $\frac{7}{9}$
i.e, $\delta ={180}^o$ or ${\cos }^{-1}\left(\frac{7}{9}\right)$
If we put ${180}^o$, we find that $v_1$ and $v_2$ are of opposite signs.
Hence $\delta ={180}^o$ is not applicable.
$\therefore \delta ={\cos }^{-1}\left(\frac{7}{9}\right)$