Question

Support of a simple pendulum of length $l$ is being accelerated in a plane at an angle of ${30}^{\circ }$ with the horizontal as shown in the diagram with an acceleration of magnitude $g$ (earth’s acceleration due to gravity). Find the new time period of the oscillation of simple pendulum.

• A. $2\pi \sqrt{\frac{l}{3g}}$
• B. $2\pi \sqrt{\frac{l}{\sqrt{3}g}}$
• C. $2\pi \sqrt{\frac{l}{g}}$
• D. $2\pi \sqrt{\frac{3l}{g}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{l}{\sqrt{3}g}}$

The time period of simple pendulum is given by
$T=2\pi \sqrt{\frac{l}{g_{eff}}}$
Pseudo force acting on the pendulum is in the direction opposite to the acceleration.

Net acceleration, $g_{eff}=\sqrt{g^2+g^2+2g^2\cos {60}^{\circ }}$

$\Rightarrow g_{eff}=\sqrt{3g^2}$

$\Rightarrow g_{eff}=\sqrt{3}g$

$\Rightarrow T=2\pi \sqrt{\frac{l}{\sqrt{3}g}}$

Hence, option $\left(b\right)$ is correct.