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 :
- $\tan^{-1}(\sqrt{2})$
- $2\tan^{-1}\left(\frac{1}{2}\right)$
- $\tan^{-1}\left(\frac{1}{2}\right)$
- $2\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to analyze the motion of a pendulum in a vertical plane. The pendulum is described as having equal magnitude of acceleration in its extreme and lowest positions.
- Let \(l\) be the length of the pendulum and \(\theta\) be the angle of skew in the extreme position.
- In the extreme position, the only acceleration is centripetal, which is due to gravity acting along the tangent of the path. Hence, the centripetal acceleration is just the gravitational component along the arc, \(g \sin \theta\).
- At the lowest position, the acceleration is due to the change in velocity as the pendulum swings. Here, the net acceleration arises because of tension offsetting gravitational acceleration.
- Given that the magnitude of the acceleration in both positions is equal, we equate:
- \(g \sin \theta = \frac{v^2}{l}\) at the lowest position.
- We know the energy conservation principle for the pendulum says:
- \(mgl(1-\cos\theta) = \frac{1}{2}mv^2\) (potential energy at the extreme position converts into kinetic energy at the lowest point).
- Solve for \(v^2\):
- \(v^2 = 2gl(1-\cos\theta)\).
- Substitute \(v^2\) back into the centripetal equation:
- \(g \sin \theta = \frac{2gl(1-\cos\theta)}{l}\)
- This simplifies to:
- \(\sin \theta = 2(1-\cos\theta)\)
- This trigonometric equation simplifies to:
- \(\tan(\theta/2) = \frac{1}{2}\)
- Thus, \(\theta = 2\tan^{-1}\left(\frac{1}{2}\right)\).
The correct option is \(2\tan^{-1}\left(\frac{1}{2}\right)\).
Approach Solution -2
Loss in kinetic energy equals gain in potential energy:
\(\frac{1}{2}mv^2 = mg\ell(1 - \cos \theta).\)
From the equation:
\(v^2 = 2g\ell(1 - \cos \theta).\)
Acceleration at the lowest point is given by:
\(\text{Acceleration} = \frac{v^2}{\ell} = 2g(1 - \cos \theta).\)
Acceleration at the extreme point:
\(a = g \sin \theta.\)
Equating the magnitudes of acceleration:
\(2g(1 - \cos \theta) = g \sin \theta \implies \sin \theta = 2(1 - \cos \theta).\)
Simplifying gives:
\(\theta = 2 \tan^{-1} \left(\frac{1}{2}\right).\)
The Correct answer is: $2\tan^{-1}\left(\frac{1}{2}\right)$
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