The time period of oscillation of a simple pe | vedclass

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(A) When a vehicle moves down a frictionless inclined plane of inclination $\alpha$,its acceleration is $a = g \sin \alpha$ directed down the plane.To

Vedclass · Accepted Answer

$2 \pi \sqrt{L / (g \cos \alpha)}$

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(A) When a vehicle moves down a frictionless inclined plane of inclination $\alpha$,its acceleration is $a = g \sin \alpha$ directed down the plane.To find the effective acceleration due to gravity $(g_{eff})$ inside the vehicle,we consider the pseudo-force acting on the pendulum bob.The effective acceleration is the vector sum of the acceleration due to gravity $(g)$ and the negative of the vehicle's acceleration $(-a)$.$g_{eff} = \vec{g} - \vec{a}$.Resolving these into components perpendicular to the inclined plane,we find $g_{eff} = g \cos \alpha$.The time period of a simple pendulum is given by $T = 2 \pi \sqrt{L / g_{eff}}$.Substituting $g_{eff} = g \cos \alpha$,we get $T = 2 \pi \sqrt{\frac{L}{g \cos \alpha}}$.

The time 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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