Question

A simple pendulum is released from re stat the horizontally stretched position. When the string makes an angle $\theta$ with the vertical, the angle which the acceleration vector of the bob makes with the string is given by

• A. $\varphi =0$
• B. $\varphi ={\tan }^{-1}\left(\frac{\tan \theta }{2}\right)$
• C. $\varphi ={\tan }^{-1}\left(2\tan \theta \right)$
• D. $\varphi =\frac{\pi }{2}$

Answer 1

Answer: (B)

Let $v$ is the velocity of bob at position $\theta$ after being released from $\theta$ horizontal position

From energy conservation, we have
$mgh=\frac{1}{2}m{v}^2$
$\Rightarrow mg(l\cos \theta )=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{v^2}{l}=2g\cos \theta ...(i)$
Restoring force on bob is component of weight.

So, tangential component of acceleration is
$a_t=g\sin \theta \dots \left(ii\right)$
If total acceleration a makes angle $\varphi$ with string. Then,

Tangential component
$a_t=a\sin \varphi$
and radial component, $a_c=a\cos \varphi$
So, $\frac{a_t}{a_c}=\frac{a\sin \varphi }{a\cos \varphi }=\tan \varphi$
$\Rightarrow \tan \varphi \frac{a_t}{a_c}\dots \left(iii\right)$
Substituting values of $a_t$ and $a_c$ from Eqs.
(i) and (ii) in Eq. (iii), we get
$\tan \varphi =\frac{g\sin \theta }{2g\cos \theta }$
$\Rightarrow \tan \varphi =\frac{\tan \theta }{2}$
so, $\varphi ={\tan }^{-1}\left(\frac{\tan \theta }{2}\right)$