JEE PYQ: Oscillations Question 1
Question 1 - 2021 (16 Mar 2021 Shift 1)
Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g/2$, the time period of pendulum will be:
(1) $\sqrt{3}, T$
(2) $\frac{T}{\sqrt{3}}$
(3) $\sqrt{\frac{3}{2}}, T$
(4) $\sqrt{\frac{2}{3}}, T$
Show Answer
Answer: (4)
Solution
When lift is stationary: $T = 2\pi\sqrt{\frac{L}{g}}$. When lift moves upwards, $g_{\text{eff}} = g + \frac{g}{2} = \frac{3g}{2}$. New time period $T’ = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} = 2\pi\sqrt{\frac{2L}{3g}} = \sqrt{\frac{2}{3}},T$.
BETA
