Question

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a,$ then the time period is given by $r=2\pi \sqrt{\frac{l}{g^{\prime }}}$ where $g^{\prime }$ is equal to

• A. $g$
• B. $g-a$
• C. $\sqrt{g^2+a^2}$
• D. $g+a$

Answer 1

The correct option is C $\sqrt{g^2+a^2}$

The pendulum is suspended from the roof of a trolley which is moving in horizontal direction with acceleration$a$

Hence the force acting will be

$F=mai-mgj$

Let $\theta$ angle between the horizontal and vertical then

$\tan \theta =\frac{ma}{mg}$

Then the value of $g^{\prime }=\sqrt{g^2+a^2}$