JEE Mains · Physics · STD 11 - 13. oscillations
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)
- A \(\sqrt{\frac{6}{5}} T\)
- B \(\sqrt{\frac{5}{6}} T\)
- C \(\sqrt{\frac{6}{7}} T\)
- D \(\sqrt{\frac{7}{6}} T\)
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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