A) T/4
B) \[\left( \frac{\sqrt{3}}{4} \right)T\]
C) \[\left( \frac{\sqrt{3}}{2} \right)T\]
D) \[\frac{T}{2}\]
Correct Answer: C
Solution :
The periodic time of pendulum in SHM is \[T=2\pi \sqrt{\frac{l}{g}}\] ?(i) where, \[l\] is length of pendulum and g is gravity. Effective gravity when lift accelerates upwards is \[g'=g+\frac{g}{3}=\frac{4}{3}g\] \[\therefore \] \[T'=2\pi \sqrt{\frac{l}{(4/3)g}}\] ?(ii) Dividing Eq. (i) by (ii),we get \[\frac{T}{T'}=\sqrt{\frac{\frac{4}{3}g}{g}}=\sqrt{\frac{4}{3}}\] \[\therefore \] \[T'=\sqrt{\frac{3}{4}}T=\frac{\sqrt{3}}{2}T\]You need to login to perform this action.
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