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13.Oscillations
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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
$\sqrt{\frac{7}{6}} T$
(JEE MAIN-2022)
Solution
$T =2 \pi \sqrt{\frac{\ell}{ g _{\text {eff }}}}$
$(a)$ when $a =0, T =2 \pi \sqrt{\frac{\ell}{ g }}$
$(b)$ when $a =\frac{ g }{6}, T ^{\prime}=2 \pi \sqrt{\frac{\ell}{ g +\frac{ g }{6}}}$
$\therefore T ^{\prime}=\sqrt{\frac{6}{7}} T$
Standard 11
Physics
