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  • The time period of oscillation of a simple pendulum of length suspended from the roof of a vehicle, w

The time period of oscillation of a simple pendulum of length \mathrm{L} suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination \alpha, is given by :
 

Option: 1

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


Option: 2

2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \sin \alpha)}
 


Option: 3

2 \pi \sqrt{\mathrm{L} / \mathrm{g}}


Option: 4

2 \pi \sqrt{\mathrm{L} /(\mathrm{g} \tan \alpha)}


Answers (1)

best_answer

\mathrm{g_{\text {eff }} =\sqrt{g^2+(g \sin \alpha)^2+2 g \times g \sin \alpha \cos (90+\alpha)} }

        \mathrm{=g \cos (\alpha ) }

\mathrm{T=2 \pi \sqrt{\frac{\ell}{g_{eff}}}=2 \pi \sqrt{\frac{\ell}{g \cos \alpha}}}

Hence 1 is correct option.





 

Posted by

Suraj Bhandari

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