Question

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 :

• A. $\tan^{-1}(\sqrt{2})$
• B. $2\tan^{-1}\left(\frac{1}{2}\right)$
• C. $\tan^{-1}\left(\frac{1}{2}\right)$
• D. $2\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Answer 1

The Correct Option is B

To address this problem, an analysis of pendulum motion in a vertical plane is required. The pendulum is characterized by identical magnitudes of acceleration at its extreme and lowest points.

1. Define the pendulum length as \(l\) and the extreme position's angle of skew as \(\theta\).
2. At the extreme position, acceleration is solely centripetal, driven by gravity along the path's tangent. Therefore, the centripetal acceleration equals the gravitational component along the arc, \(g \sin \theta\).
3. At the lowest position, acceleration results from the velocity change during the pendulum's swing. Here, the net acceleration occurs as tension counteracts gravitational acceleration.
4. Equate the magnitudes of acceleration at both positions, as they are given to be equal:
5. \(g \sin \theta = \frac{v^2}{l}\) at the lowest position.
6. The principle of energy conservation for the pendulum states:
7. \(mgl(1-\cos\theta) = \frac{1}{2}mv^2\) (potential energy at the extreme position transforms into kinetic energy at the lowest point).
8. Rearrange to solve for \(v^2\):
9. \(v^2 = 2gl(1-\cos\theta)\).
10. Substitute this expression for \(v^2\) into the centripetal acceleration equation:
11. \(g \sin \theta = \frac{2gl(1-\cos\theta)}{l}\)
12. This equation simplifies to:
13. \(\sin \theta = 2(1-\cos\theta)\)
14. Further simplification of this trigonometric equation yields:
15. \(\tan(\theta/2) = \frac{1}{2}\)
16. Consequently, \(\theta = 2\tan^{-1}\left(\frac{1}{2}\right)\).

The correct selection is \(2\tan^{-1}\left(\frac{1}{2}\right)\).