WBJEE · Physics · Oscillations
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
- A \(2 \pi \sqrt{\frac{1}{g \cos \alpha}}\)
- B \(2 \pi \sqrt{\frac{1}{g \sin \alpha}}\)
- C \(2 \pi \sqrt{\frac{1}{g}}\)
- D \(2 \pi \sqrt{\frac{1}{\mathrm{~g} \tan \alpha}}\)
Answer & Solution
Correct Answer
(A) \(2 \pi \sqrt{\frac{1}{g \cos \alpha}}\)
Step-by-step Solution
Detailed explanation
Hints: \(\begin{aligned} & \mathrm{g}_{\text {eff }}=\mathrm{g} \cos \alpha \\ & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\text {eff }}}} \end{aligned}\)
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