A simple pendulum has a time period T when measured in a stationary elevator. If the elevator begins to accelerate upwards at a rate of g/2, what is the new time period of the pendulum?

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EEJ MAIN Physics QUESTION #7076

Question 1

A simple pendulum has a time period $T$ when measured in a stationary elevator. If the elevator begins to accelerate upwards at a rate of $g/2$, what is the new time period of the pendulum?

$\sqrt{\frac{2}{3}}T$✔️
$\sqrt{3}T$
$\sqrt{\frac{3}{2}}T$
$\frac{T}{\sqrt{3}}$

Correct Answer Explanation

The effective gravity $g_{eff}$ in an upward accelerating lift is $g + a = g + g/2 = 3g/2$. Since the time period $T \propto \frac{1}{\sqrt{g}}$, the new period $T' = T \sqrt{\frac{g}{3g/2}} = T \sqrt{\frac{2}{3}}$.

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