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
$\Rightarrow \frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} \ell(1-\cos \theta)$
$\Rightarrow \frac{\mathrm{v}^{2}}{\ell}=2 \mathrm{~g}(1-\cos \theta)$
Acceleration at lowest point $=\frac{\mathrm{v}^{2}}{\ell}$
Acceleration at extreme point $=g \sin \theta$
Hence, $\frac{\mathrm{v}^{2}}{\ell}=\mathrm{g} \sin \theta$
$\therefore \sin \theta=2(1-\cos \theta)$
$\Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)$
