Question

Two particles $P_1$ and $P_2$ are performing $\text{SHM}$ 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 $12\text{sec}$ and the difference of their amplitude is $12\text{cm}$, then find the minimum time in $\text{seconds}$ after which the separation between the particles become $6\text{cm}$.

• A. $5\text{s}$
• B. $4\text{s}$
• C. $2\text{s}$
• D. $8\text{s}$

Answer 1

The correct option is C $2\text{s}$

The equation representing position for the two particles starting from their positive extreme position at $t=0$ is given as,
$x_1=A_1\cos \omega t....\left(i\right)$
$x_2=A_2\cos \omega t....\left(ii\right)$
Since time period for both particles is same, their angular frequency will be same i.e ${\omega }_1={\omega }_2=\omega$

On substracting $\left(i\right)$ and $\left(ii\right)$,
$x_1-x_2=(A_1-A_2)\cos \omega t=12\cos \omega t....\left(iii\right)$

$(\because A_1-A_2=12\text{cm})$, given the difference of their amplitudes

Let at time $t$ the separation between particles become $6\text{cm}$
$\Rightarrow x_1-x_2=6$
substituting from Eq.$\left(iii\right)$
$\Rightarrow 12\cos \omega t=6$
$\Rightarrow \cos \omega t=\frac{1}{2}$
$\Rightarrow \omega t=\frac{\pi }{3}$
$\Rightarrow \frac{2\pi }{T}.t=\frac{\pi }{3}$
$(\because T=12\text{sec})$
$\Rightarrow \frac{2\pi }{12}.t=\frac{\pi }{3}$
$\therefore t=2\text{s}$