Question

A pendulum is oscillating with an angular amplitude of $90^\circ $ as shown in the figure. The value of angle $\alpha $ , made by the string with the vertical, at which the acceleration is directed horizontally is

• A. 0
• B. $90^\circ $
• C. $\cos^{- 1} \left(\frac{1}{\sqrt{3}}\right)$
• D. $\sin^{- 1} \left(\frac{1}{\sqrt{3}}\right)$

Answer 1

Answer: (C)

For horizontal acceleration, net force in vertical direction should be zero.

Loss in potential energy = gain in kinetic energy
$mgh=\frac{1}{2}mv^{2}$
$mgl\cos \alpha =\frac{1}{2}mv^{2}$
$\therefore v=\sqrt{2 g l \cos \alpha }$
$\because $ The net force towards O is
$T-mg\cos \alpha =\frac{m v^{2}}{l}$
$=\frac{m 2 g l \cos \alpha \, }{l}=2 \, mg\cos⁡\alpha $
$\therefore T=3mg\cos \alpha $
$\therefore T\cos \alpha =mg$
Or $3\,mg \, \cos^{2} \, \alpha =mg$
$\therefore \alpha =\left(\cos\right)^{- 1}\left(\frac{1}{\sqrt{3}}\right)$