Question 118 of 194

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 :

A$2\pi \sqrt{\dfrac{L}{\left(g\cos \alpha \right)}}$B$2\pi \sqrt{\dfrac{L}{\left(g\sin \alpha \right)}}$C$2\pi \sqrt{\dfrac{L}{g}}$D$2\pi \sqrt{\dfrac{L}{\left(g\tan \alpha \right)}}$

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Answer: (A) $2\pi \sqrt{\dfrac{L}{\left(g\cos \alpha \right)}}$

Acceleration of vehicle down the incline will be $a=g\sin \alpha$
In the frame of the vehicle, we have to apply pseudo force. The Force diagram is shown in figure.
Component of $mg$ along the inclined (i.e. $mg\sin \alpha$) will balance the pseudo force. Therefore, ${g}_{eff}=gcos\alpha$


Hence,
$T=2\pi \sqrt{\dfrac{L}{{g}_{eff}}}=2\pi \sqrt{\dfrac{L}{gcos\alpha }}$.