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. ${\left(cos\right)}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. ${\left(sin\right)}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 1

Answer: (C)

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

Loss in potential energy = gain in kinetic energy
$mgh=\frac{1}{2}m{v}^2$
$mglcos\alpha =\frac{1}{2}m{v}^2$
$\therefore v=\sqrt{2glcos\alpha }$
$\because$ The net force towards O is
$T-mgcos\alpha =\frac{m{v}^2}{l}$
$=\frac{m2glcos\alpha }{l}=2mgcos\alpha$
$\therefore T=3mgcos\alpha$
$\therefore Tcos\alpha =mg$
Or $3mgco{s}^2\alpha =mg$
$\therefore \alpha ={\left(cos\right)}^{-1}\left(\frac{1}{\sqrt{3}}\right)$