### Oscillations Question 3

#### Question 3 - 2024 (27 Jan Shift 2)

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :

(1) $\tan ^{-1}(\sqrt{2})$

(2) $2 \tan ^{-1}\left(\frac{1}{2}\right)$

(3) $\tan ^{-1}\left(\frac{1}{2}\right)$

(4) $2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)$

## Show Answer

#### Answer: (2)

#### Solution:

Loss in kinetic energy = Gain in potential energy

$\Rightarrow \frac{1}{2} mv^{2}=mg \ell(1-\cos \theta)$

$\Rightarrow \frac{v^{2}}{\ell}=2 g(1-\cos \theta)$

Acceleration at lowest point $=\frac{v^{2}}{\ell}$

Acceleration at extreme point $=g \sin \theta$

Hence, $\frac{v^{2}}{\ell}=g \sin \theta$

$\therefore \sin \theta=2(1-\cos \theta)$

$\Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)$