Thursday, 13 June 2019

homework and exercises - Period of pendulum in falling lift


A simple pendulum suspended from the ceiling of a stationary lift has period $T_0$. When the lift descends at steady speed, the period is $T_1$. When the lift descends with constant downward acceleration, the period is $T_2$. Explain why $T_0 = T_1 < T_2$.


I know that for small angles of oscillations, $T \approx 2\pi\sqrt\frac{l}{g}$ where $l$ is the length of the pendulum. For the first two cases, since $\sum F = 0$, the value of $g$ is apparently the same. However, I am unable to explain why the period is largest for $T_2$. If $g$ is the same for all cases, shouldn't their periods be the same?




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