A simple pendulum is oscillating with an angular amplitude of 90^o as shown in fig. (4). The value of θ for

which the resultant acceleration of the bob is directed horizontally is

See fig. (11).

From figure     ${\mathrm{a}}_{\mathrm{t}}=\mathrm{gsin}\mathrm{\theta }$

and    ${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^2}{\mathrm{l}}=\frac{2\mathrm{glcos}\mathrm{\theta }}{\mathrm{l}}=2\mathrm{gcos}\mathrm{\theta }$

Now,    $\tan \mathrm{\theta }=\frac{{\mathrm{a}}_{\mathrm{c}}\sin {90}^{\circ }}{{\mathrm{a}}_{\mathrm{t}}+{\mathrm{a}}_{\mathrm{c}}\cos {90}^{\circ }}$

or       $\tan \mathrm{\theta }=\frac{2\mathrm{gcos}\mathrm{\theta }}{\mathrm{gsin}\mathrm{\theta }}=\frac{2}{\tan \mathrm{\theta }}$

or      ${\tan }^2\mathrm{\theta }=2\text{ or }{\sec }^2\mathrm{\theta }=3$

$\therefore \cos \mathrm{\theta }=1/\sqrt{3}$

$\mathrm{\theta }={\cos }^{-1}(1/\sqrt{3})$