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Question: Answered & Verified by Expert
A simple pendulum of time period $1 \mathrm{~s}$ and length $l$ is hung from a fixed support at $O$. Such that the bob is at a distance $H$ vertically above $A$ on the ground (figure) the amplitude is $\theta_0$ the string snaps at $\theta=\theta_0 / 2$. Find the time taken by the bob to hit the ground. Also find distance from $A$ where bob hits the ground. Assume $\theta_0$ to be small, so that $\sin \theta_0 \simeq \theta_0$ and $\cos \theta_0 \simeq 1$

PhysicsOscillations
Solution:
1510 Upvotes Verified Answer
Consider the given figure at $t=0, \theta=\frac{\theta}{2}$


Let us consider $t=0$ when $\theta=\theta_0$, then $\theta=\theta_0 \cos \omega t$ is given that a seconds pendulum
$$
\omega=2 \pi \Rightarrow \theta=\theta_0 \cos 2 \pi t
$$
At time $t_1$, let $\theta=\theta_0 / 2$
$$
\begin{aligned}
&\text { so } \frac{\theta_0}{2}=\theta_0 \cos \omega t_1 \quad \text { from (i) } \\
&\therefore \cos 2 \pi t_1=1 / 2 \\
&{\left[\because \cos 2 \pi t_1=\cos \frac{\pi}{3} \Rightarrow 2 \pi t_1=\frac{\pi}{3}\right]} \\
&2 \pi t_1=\frac{\pi}{3} \text { or } t_1=\frac{1}{6} \quad \text { [from Eq. (i)] }
\end{aligned}
$$
$$
\frac{d \theta}{d t}=-\left(\theta_0 2 \pi\right) \sin 2 \pi t
$$
At $\mathrm{t}=\mathrm{t}_1=\frac{1}{6}$ i.e. at $\left(\theta=\frac{\theta_0}{2}\right)$.
$$
\frac{d \theta}{d t}=-\theta_0 2 \pi \sin \frac{2 \pi}{6}=-\sqrt{3} \pi \theta_0
$$
So, the linear velocity is
$u=-\sqrt{3} \pi \theta_0 l$ perpendicular to the string and (-ve) sign shows the bob's motion is towards left.
$$
u_y=u \sin \left(\frac{\theta_0}{2}\right)=-\sqrt{3} \pi \theta_0 l \sin \left(\frac{\theta_0}{2}\right)
$$
and the horizontal component is
$$
u_x=u \cos \left(\frac{\theta_0}{2}\right)=-\sqrt{3} \pi \theta_0 l \cos \left(\frac{\theta_0}{2}\right)
$$
At the time it snaps, the vertical height is
$$
H^{\prime}=H+l\left(1-\cos \left(\frac{\theta_0}{2}\right)\right)
$$
Let vertical distance covered by $\left(u_y\right)$ is $H$ and the time required for fall be $t$, then
$$
\begin{aligned}
&s=u t+\frac{1}{2} g t^2 \\
&H^{\prime}=u_y t+\left(\frac{1}{2}\right) g t^2
\end{aligned}
$$
(notice $g$ also in the negative direction)
$$
\begin{aligned}
&\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \sin \left(\frac{\theta_0}{2}\right) \times t-H^{\prime}=0 \\
&\sin \left(\frac{\theta_0}{2}\right) \simeq\left(\frac{\theta_0}{2}\right) \\
&\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \frac{\theta_0}{2} t-H^{\prime}=0
\end{aligned}
$$
or $\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \sin \frac{\theta_0}{2} t-H^{\prime}=0$
Above equation is quadratic, hence
$$
\therefore t=\frac{-\sqrt{3} \pi \theta_0 l \sin \frac{\theta_0}{2} \pm \sqrt{3 \pi^2 \theta_0^2 l^2 \sin ^2 \frac{\theta_0}{2}+2 g H^{\prime}}}{g}
$$
$$
t=\frac{-\sqrt{3} \pi l \frac{\theta_0^2}{2} \pm \sqrt{3 \pi^2\left(\frac{\theta_0^4}{4}\right) l^2+2 g H^{\prime}}}{g}[
$$
As given that $\theta_0$ is small, hence neglecting terms of higher order i.e. $\theta_0^2$ and $\theta_0^4$
$$
t=\sqrt{\frac{2 H^{\prime}}{g}}
$$
[from Eq. (iii)]
$$
\begin{aligned}
&\text { Now, } H^{\prime}=H+l(1-1) \quad\left[\because \cos \theta_0 / 2=1\right] \\
&=H \\
&t=\sqrt{\frac{2 H}{g}} \quad\left(\because H=H^{\prime}\right) \\
&
\end{aligned}
$$
The distance covered in the horizontal direction is $u_x t$ to the left of where the bob is snapped
$$
X=U x t=\left[\sqrt{3} \pi \theta_0 I \cos \left(\frac{\theta_0}{2}\right) \sqrt{\frac{2 H}{g} s}\right]
$$
As given that, $\theta_0$ is small so $\cos \left(\frac{\theta_0}{2}\right)=1$
$$
X=\sqrt{3} \pi \theta_0 l \sqrt{\frac{2 H}{g}}=\sqrt{\frac{6 H}{g}} \theta_0 l \pi
$$
At the time of snapping, the bob was at a distance of $l \sin \left(\frac{\theta_0}{2}\right)=l \frac{\theta_0}{2}$ from $A$.

Thus, the distance of bob from $A$ where it meets the ground is
$$
\begin{aligned}
&=\frac{l \theta_0}{2}-X=\frac{l \theta_0}{2}-\sqrt{\frac{6 H}{g}} \theta_0 l \pi \\
&=\theta_0 l\left(\frac{1}{2}-\pi \sqrt{\frac{6 H}{g}}\right)
\end{aligned}
$$

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