The period of oscillation of a simple pendulu | vedclass

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(A) When a vehicle moves down an inclined plane of inclination $\alpha$ without friction, it experiences an acceleration $a = g \sin \alpha$ down the

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{l}{g \cos \alpha}}$

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(A) When a vehicle moves down an inclined plane of inclination $\alpha$ without friction, it experiences an acceleration $a = g \sin \alpha$ down the plane.To find the effective acceleration $g_{	ext{eff}}$ experienced by the pendulum inside the vehicle, we consider the frame of reference of the vehicle.In this non-inertial frame, a pseudo-force $ma$ acts on the pendulum bob in the upward direction along the incline.The effective acceleration $g_{	ext{eff}}$ is the vector sum of the gravitational acceleration $\vec{g}$ and the pseudo-acceleration $-\vec{a}$.Resolving $\vec{g}$ into components, we have $g \cos \alpha$ perpendicular to the incline and $g \sin \alpha$ parallel to the incline.The pseudo-acceleration is $a = g \sin \alpha$ directed up the incline.Thus, the components parallel to the incline cancel out $(g \sin \alpha - g \sin \alpha = 0)$, and the effective acceleration is $g_{	ext{eff}} = g \cos \alpha$ directed perpendicular to the incline.The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{	ext{eff}}}}$.Substituting $g_{	ext{eff}} = g \cos \alpha$, we get $T = 2 \pi \sqrt{\frac{l}{g \cos \alpha}}$.

The period of oscillation of a simple pendulum of length $l$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by

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