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 position. If the time period of each particle is 12 s and the difference of their amplitudes is 12 cm then find the minimum time (in seconds) after which the separation between the particles become 6 cm.

Answer 1

The $\text{x}$ coordinates of the particles are
$x_{1}=A_{1}cos \omega t , \, x_{2} = A_{2}cos ⁡ \omega t$
Separation $x_{1}-x_{2}=\left(A_{1} - A_{2}\right)cos \omega t=12cos ⁡ \omega t$
Now $x_{1}-x_{2}=6=12cos \omega t$
$\Rightarrow \omega t=\frac{\pi }{3}\Rightarrow \frac{2 \pi }{12}\times t=\frac{\pi }{3}$
$\Rightarrow t=2 \, s$