Question

Two particles $P_1$ and $P_2$ are performing SHM along the same line about the same mean position. Initially they are at their positive extreme positions. If the time period of each particle is $12sec$ and the difference of their amplitudes is $12cm$ then find the minimum time after which the separation between the particles become $6cm$.

Answer 1

If they are at their positive extreme positions, then $Asin\omega t$ will become $Acos\omega t$

Let $x_2$ be the one with the lesser amplitude. It is less than that of $x_1$ by 12cm. So $⟶$

$x_{{}_1}-x_{{}_2}=6$

$x_{{}_1}=Acos\omega t$

$x_{{}_1}=(A-12)cos\omega t$

Substituting in 1st equation, we get

$Acos\omega t-(A-12)cos\omega t=12cos\omega t$

Since T=12sec and $\omega =\frac{2\pi }{T},$ we get $⟶$

$12cos\omega t=12cos\left(\frac{\pi }{6}\right)t=6$

$6\sqrt{3}t=6$------------------------------------------------$cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}$

$t=\frac{1}{\sqrt{3}}sec=0.5773sec\approx 6sec$