Question

Two particles move parallel to $x-$ axis about the origin with the same amplitude and frequency. At a certain instant, they are found at distance $\frac{A}{3}$ from the origin on opposite sides, but their velocities are found to be in the same direction. Find the phase difference between the two particles.

• A. $\pi$
• B. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• C. $\frac{\pi }{3}$
• D. ${\cos }^{-1}\left(\frac{7}{9}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{7}{9}\right)$

Let equations of the two SHMs be
$x_1=A\sin \omega t......\left(1\right)$
$x_2=A\sin (\omega t+\varphi ).......\left(2\right)$
At time $t$, let $x_1=\frac{A}{3},x_2=-\frac{A}{3}$
$\Rightarrow \frac{A}{3}=A\sin \omega t$
& $-\frac{A}{3}=A\sin (\omega t+\varphi )$
which gives, $\sin \omega t=\frac{1}{3}......\left(3\right)$
& $\sin (\omega t+\varphi )=-\frac{1}{3}.....\left(4\right)$
From Eq. $\left(3\right)$ and $\left(4\right)$, using $\sin (A+B)=\sin A\cos B+\cos A\sin B$, we can write that
$\sin \omega t\cos \varphi +\cos \omega t\sin \varphi =-\frac{1}{3}$
$\Rightarrow \frac{1}{3}\cos \varphi +\sqrt{1-\frac{1}{9}}\sin \varphi =-\frac{1}{3}$
$\Rightarrow \sqrt{\frac{8}{9}}\sin \varphi =-\frac{1}{3}-\frac{1}{3}\cos \varphi$
Squaring on both sides, we get
$\left(\frac{8}{9}\right)(1-{\cos }^2\varphi )=\frac{1}{9}(1+{\cos }^2\varphi +2\cos \varphi )$
$\Rightarrow {\cos }^2\varphi +2\cos \varphi +1+8{\cos }^2\varphi -8=0$
$\Rightarrow 9{\cos }^2\varphi +2\cos \varphi -7=0$
$\Rightarrow 9{\cos }^2\varphi +9\cos \varphi -7\cos \varphi -7=0$
$\Rightarrow \cos \varphi =\left(\frac{7}{9}\right)$ or $-1$
$\Rightarrow \varphi =\pi$ or ${\cos }^{-1}\left(\frac{7}{9}\right)$
Differentiating Eqs. $\left(1\right)$ and $\left(2\right)$, we obtain
$v_1=A\omega \cos \omega t....\left(5\right)$
and $v_2=A\omega \cos (\omega t+\varphi )......\left(6\right)$
If we put $\varphi =\pi$, we find $v_1$ and $v_2$ are of opposite signs.
Hence, $\varphi =\pi$ is not acceptable.
$\therefore$ Phase difference between the two particles is $\varphi ={\cos }^{-1}\left(\frac{7}{9}\right)$
Thus, option (d) is the correct answer.