JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2022
- Medium
- MCQ
Time period of a simple pendulum in a stationary lift is ' \(T\) '. If the lift accelerates with \(\frac{ g }{6}\) vertically upwards then the time period will be ..... (where \(g =\) acceleration due to gravity)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{6}{7}} T\)
Step-by-step Solution
Detailed explanation
\(T =2 \pi \sqrt{\frac{\ell}{ g _{\text {eff }}}}\)
\((a)\) when \(a =0, T =2 \pi \sqrt{\frac{\ell}{ g }}\)
\((b)\) when \(a =\frac{ g }{6}, T ^{\prime}=2 \pi \sqrt{\frac{\ell}{ g +\frac{ g }{6}}}\)
\(\therefore T ^{\prime}=\sqrt{\frac{6}{7}} T\)
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