Question

A simple pendulum is oscillating with an angular amplitude ${90}^0$. If the direction of resultant acceleration of the bob is horizontal at a point where angle made by the string with vertical is:

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

Answer 1

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

the resultant acceleration
of bob is horizontal, that is
verticle force must be equal.
$Tcos\theta =mg{a}_T=gsin\theta$
$a_C=\frac{v^2}{R}$
Using WET
$H=Rcos\theta$
$mgRcos\theta =\frac{m}{2}{v}^2\Rightarrow V=\sqrt{2gRcos\theta }$
$\therefore a_c=2gcos\theta$
$VerticalAccelerationisZero$
$\left(2gcos\theta \right)cos\theta =\left(gsin\theta \right)sin\theta$
$tan\theta =\frac{1}{\sqrt{2}}$
$\therefore cos\theta =\frac{1}{\sqrt{3}}$