Question

Two simple pendulums $A$ and $B$ having length $l$ and $\frac{l}{4}$ respectively are released from the positions shown in the figure. If the time after release when the two strings become parallel for the first time is given by $\frac{2\pi }{x}\sqrt{\frac{l}{g}}\text{s}$, then the value of $x$ will be
[Assume angle $\theta$ is very small and motion starts at $t=0$]

Answer 1

Let at any time $t$ after release, the angular position of the strings be ${\theta }_1$ and ${\theta }_2$ $w.r.t$ $X{X}^{\prime }$.


We know that, angular frequency
$\omega =\sqrt{\frac{g}{l}}$
So, ${\omega }_1=\sqrt{\frac{g}{l/4}}=2\sqrt{\frac{g}{l}}$ and ${\omega }_2=\sqrt{\frac{g}{l}}$
Now, ${\theta }_1=\theta \cos \left({\omega }_{1}t\right)$
$\Rightarrow {\theta }_1=\theta \cos \left(2\sqrt{\frac{g}{l}}t\right)$ .......(1)
And, ${\theta }_2=-\theta \cos \left({\omega }_{2}t\right)$
$\Rightarrow {\theta }_2=\theta \cos (-\sqrt{\frac{g}{l}}t)$ ..........(2)
For the strings to become parallel, the displacement angles should be the same.
So, $\cos \left(2\sqrt{\frac{g}{l}}t\right)=\cos (-\sqrt{\frac{g}{l}}t)$
[from (1) and (2)]
$\Rightarrow 2\sqrt{\frac{g}{l}}t=2n\pi \pm \sqrt{\frac{g}{l}}t$
For $n=0,t=0$ [trivial]
For $n=1,t=\frac{2\pi }{3}\sqrt{\frac{l}{g}}$(minimum) and $t=2\pi \sqrt{\frac{l}{g}}$
So, after $t=\frac{2\pi }{3}\sqrt{\frac{l}{g}}$, two strings will become parallel for the first time.
$t=\frac{2\pi }{x}\sqrt{\frac{l}{g}}$ [given]
Hence, $x=3$