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Physics Question on simple harmonic motion

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 θ\theta of thread deflection in the extreme position will be :

A

tan⁡−1(2)\tan^{-1}(\sqrt{2})

B

2tan⁡−1(12)2\tan^{-1}\left(\frac{1}{2}\right)

C

tan⁡−1(12)\tan^{-1}\left(\frac{1}{2}\right)

D

2tan⁡−1(15)2\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)

Answer

2tan⁡−1(12)2\tan^{-1}\left(\frac{1}{2}\right)

Explanation

Solution

Loss in kinetic energy equals gain in potential energy:

12mv2=mgℓ(1−cos⁡θ).\frac{1}{2}mv^2 = mg\ell(1 - \cos \theta).

From the equation:

v2=2gℓ(1−cos⁡θ).v^2 = 2g\ell(1 - \cos \theta).

Acceleration at the lowest point is given by:

Acceleration=v2ℓ=2g(1−cos⁡θ).\text{Acceleration} = \frac{v^2}{\ell} = 2g(1 - \cos \theta).

Acceleration at the extreme point:

a=gsin⁡θ.a = g \sin \theta.

Equating the magnitudes of acceleration:

2g(1−cos⁡θ)=gsin⁡θ  ⟹  sin⁡θ=2(1−cos⁡θ).2g(1 - \cos \theta) = g \sin \theta \implies \sin \theta = 2(1 - \cos \theta).

Simplifying gives:

θ=2tan⁡−1(12).\theta = 2 \tan^{-1} \left(\frac{1}{2}\right).

The Correct answer is: 2tan⁡−1(12)2\tan^{-1}\left(\frac{1}{2}\right)