JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 9)
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 :
$$2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \cos \alpha)}$$
$$2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \sin \alpha)}$$
$$2 \pi \sqrt{\mathrm{L} / \mathrm{g}}$$
$$2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \tan \alpha)}$$
Explanation
$$\left| {{g_{eff}}} \right| = \left| {\overline g - \overline a } \right|$$
$$ \Rightarrow {g_{eff}} = g\cos \theta $$
$$ \Rightarrow T = 2\pi \sqrt {{l \over {{g_{eff}}}}} $$
$$ = 2\pi = \sqrt {{L \over {g\cos \theta }}} $$
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