A tunnel is dug through the centre of the earth. Show that a body of mass $m$ when dropped from rest from one end of the tunnel will execute simple harmonic motion.
Consider the situation shown in the diagram.
The gravitational force on the particle at a distance $r$ from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the position of the particle. The external shells exert no force on the particle.
More clearly, let $g^{\prime}$ be the acceleration at $P$.
So, $$g^{\prime}=g\left(1-\frac{d}{R}\right)=g\left(\frac{R-d}{R}\right)$$
$$\begin{aligned} &\begin{aligned} &\text { From figure, }\quad R-d =y \\ \Rightarrow \quad g^{\prime} & =g \frac{y}{R} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Force on body at } p \text {, }\\ &F=-m g^{\prime}=\frac{-m g}{R} y\quad \text{... (i)} \end{aligned}$$
$$\begin{aligned} &\Rightarrow \quad F \propto-y \quad \text { [where, } y \text { is distance from the centre] }\\ &\text { So, motion is SHM. } \end{aligned}$$
For time period, we can write Eq. (i)
As $$m a=-\frac{M g}{R} y \Rightarrow a=-\frac{g}{R} y$$
Comparing with $a=-\omega^2 y$
$$\begin{aligned} \omega^2 & =\frac{g}{R} \\ \Rightarrow \quad\left(\frac{2 \pi}{T}\right) & =\frac{g}{R} \Rightarrow T=2 \pi \sqrt{\frac{R}{g}} \end{aligned}$$
A simple pendulum of time period $1 s$ and length $l$ is hung from a fixed support at 0 . 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 $\cos \theta_0 \simeq 1$
Consider the diagram,
Let us assume $t=0$ when $\theta=\theta_0$, then $\theta=\theta_0 \cos \omega t$
Given a seconds pendulum $\omega=2 \pi \Rightarrow \theta=\theta_0 \cos 2 \pi t\quad \text{... (i)}$
At time $t_1$ let $\theta=\theta_0 / 2$
$$\begin{aligned} \therefore \quad \cos 2 \pi t_1 & =1 / 2 \Rightarrow t_1=\frac{1}{6} \quad\left[\because \cos 2 \pi t_1=\cos \frac{\pi}{3} \Rightarrow 2 \pi t_1=\frac{\pi}{3}\right] \\ \frac{d \theta}{d t} & =-\left(\theta_0 2 \pi\right) \sin 2 \pi t \quad \text{[from Eq. (i)]} \end{aligned}$$
$$\begin{aligned} \text { At } \quad t & =t_1=\frac{1}{6} \\ \frac{d \theta}{d t} & =-\theta_0 2 \pi \sin \frac{2 \pi}{6}=-\sqrt{3} \pi \theta_0 \end{aligned}$$
Negative sign shows that it is going left.
Thus, the linear velocity is
$$u=-\sqrt{3} \pi \theta_0 l \text { perpendicular to the string. }$$
The vertical component is
$$u_y=-\sqrt{3} \pi \theta_0 l \sin \left(\theta_0 / 2\right)$$
and the horizontal component is
$$u_x=-\sqrt{3} \pi \theta_0 l \cos \left(\theta_0 / 2\right)$$
At the time it snaps, the vertical height is
$$H^{\prime}=H+l\left(1-\cos \left(\theta_0 / 2\right)\right)\quad \text{... (ii)}$$
$$\begin{aligned} &\text { Let the time required for fall be } t \text {, then }\\ &H^{\prime}=u_y t+(1 / 2) g t^2 \quad \text { (notice } g \text { also in the negative direction) } \end{aligned}$$
or $$\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \sin \frac{\theta_o}{2} t-H^{\prime}=0$$
$\therefore$ $$t=\frac{-\sqrt{3} \pi \theta_0 l \sin \frac{\theta_0}{2} \pm \sqrt{3 \pi^2 \theta_0^2 l^2 \sin ^2 \frac{\theta_0}{2}+2 g H^{\prime}}}{g}$$
$$=\frac{-\sqrt{3} \pi l \frac{\theta_0^2}{2} \pm \sqrt{3 \pi^2\left(\frac{\theta_0{ }^4}{4}\right) l^2+2 g H^{\prime}}}{g}\left[\because \sin \frac{\theta_0}{2} \approx \frac{\theta_0}{2} \text { for small angle }\right]$$
$$\begin{aligned} &\text { Given that } \theta_0 \text { is small, hence neglecting terms of order } \theta_0^2 \text { and higher }\\ &t=\sqrt{\frac{2 H^{\prime}}{g}} \quad \text{[from Eq. (iii)]} \end{aligned}$$
Now, $$H^{\prime} \approx H+l(1-1) \quad\left[\therefore \cos \theta_0 / 2 \approx 1\right]$$
$$ \begin{aligned} &\begin{aligned} & =H \quad \text { [from Eq. (ii)] }\\ \Rightarrow \quad t & =\sqrt{\frac{2 H}{g}} \end{aligned}\\ \end{aligned}$$
$$\begin{aligned} &\text { The distance travelled in the } x \text {-direction is } u_x t \text { to the left of where the bob is snapped }\\ &\begin{gathered} X=U x t=\sqrt{3} \pi \theta_0 l \operatorname{Cos}\left(\frac{\theta_0}{2}\right) \sqrt{\frac{2 H}{g}} \mathrm{~s} \\ \text { As } \theta_0 \text { is small } \Rightarrow \cos \left(\frac{\theta_0}{2}\right) \approx 1 \\ X=\sqrt{3} \pi \theta_0 l \sqrt{\frac{2 H}{g}}=\sqrt{\frac{6 H}{g}} \theta_0 l \pi \end{gathered} \end{aligned}$$
At the time of snapping, the bob was at a horizontal distance of $l \sin \left(\theta_0 / 2\right) \approx l \frac{\theta_0}{2}$ from $A$.
Thus, the distance of bob from $A$ where it meets the ground is
$$\begin{aligned} & =\frac{l \theta_0}{2}-X=\frac{l \theta_0}{2}-\sqrt{\frac{6 H}{g}} \theta_0 l \pi \\ & =\theta_0 l\left(\frac{1}{2}-\pi \sqrt{\frac{6 H}{g}}\right) \end{aligned}$$