Question

A simple pendulum is oscillating with an angular of ${90}^{\circ }$ .If the direction of the result acceleration of the bob is horizontal at a where the angle made by the string vertical is

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

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$tan\alpha =\frac{a_n}{a_t}=\frac{v^2/R}{gsin\alpha }=\frac{2gh/R}{gsin\alpha }$

$h$ is the vertical distance between fix point and bob centre.

$tan\alpha =\frac{2g\left(Rcos\alpha \right)}{Rgsin\alpha }$

$tan\alpha =\sqrt{2}$

or

$cos\alpha =\frac{1}{\sqrt{3}}$

Then,

$\alpha =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$