Question

Two particles execute SHM parallel to x-axis about the origin with the same amplitude and frequency. At a certain instant, they are found at distance A/3 from the origin on opposite sides but their velocities are found to be in the same direction. What is the phase difference between the two?

Answer 1

Let equations of two S.H.M. be
$x_1=Asin\omega t$
$x_2=Asin(\omega t+\varphi$)
Given that $(A/3=Asin\omega t)and-A/3=Asin(\omega t+\varphi$)
Which gives, sin $\omega t=1/3$
sin ($\omega t+\varphi )=-1/3$
sin $\omega$ t cos $\varphi$ + cos $\omega$ t sin $\varphi$ = -1/3
1/3 cos $\varphi$ + $\sqrt{1-1/9}$ sin $\varphi$ = -1/3
$\varphi$ = $\pi$ or $co{s}^{-1}\left(7/9\right)$
Differentiating (i) and (ii) we obtain
$v_1$ = A $\omega$ cos$\omega$t and $v_2$ = A $\omega$ cos ($\omega$ + $\varphi$)
If we put $\varphi$ = $\pi$, we find $v_1$ and $v_2$ are of opposite signs.
$\varphi$ = $co{s}_{-1}$ 7/9