Question

A simple pendulum oscillates in the plane of a smooth wedge which moves with an acceleration A = g tan $\theta$ .The period of SHM is :

• A. $2\pi \sqrt{\frac{l}{g\tan \theta }}$
• B. $2\pi \sqrt{\frac{l}{g\cos \theta }}$
• C. Zero
• D. Infinite

Answer 1

The correct option is B $2\pi \sqrt{\frac{l}{g\cos \theta }}$

We know

$T=2\pi \sqrt{\frac{l}{g_{eff}}}$

$g_{eff}=\sqrt{{\left(g\right)}^2+{\left(g\tan \theta \right)}^2}=\sqrt{{\left(g\right)}^2+{\left(g\right)}^2{\tan }^2\theta }$

$=g\sqrt{1+{\tan }^2\theta }=g\sqrt{{\sec }^2\theta }=g\cos \theta$

Hence $T=2\pi \sqrt{\frac{l}{g\cos \theta }}$