Oscillations

JEE Main level, single correct, medium

121 of 194

Question 121 of 194

Time period of a simple pendulum in a stationary lift is TT. If the lift accelerates with g6\dfrac{g}{6} vertically upwards then the time period will be
(Where g=g= acceleration due to gravity)

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Answer: (C) 67T\sqrt{\dfrac{6}{7}}T

The time period of a simple pendulum is given byT=2πlgT=2\pi \sqrt{\dfrac{l}{g}}
In an upward accelerating frame the effective geff=g+a{g}_{eff}=g+a
⇒T2=2πlg+a=2πlg+g6\Rightarrow {T}_{2}=2\pi \sqrt{\dfrac{l}{g+a}}=2\pi \sqrt{\dfrac{l}{g+\dfrac{g}{6}}}
TT2=g+g6g=76\dfrac{T}{{T}_{2}}=\sqrt{\dfrac{g+\dfrac{g}{6}}{g}}=\sqrt{\dfrac{7}{6}}
⇒T2=67T\Rightarrow {T}_{2}=\sqrt{\dfrac{6}{7}}T

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