What is the period of a 1.30-m long simple pendulum in an elevator that is moving downward with a uniform acceleration of 1.20 m/s²? A)2.29 s B)3.24 s C)2.16 s D)2.85 s E)2.44 s

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Quizplus · Accepted Answer

The period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}}$, where $L$ is the length of the pendulum and $g_{\text{eff}}$ is the effective acceleration due to gravity. In an elevator moving downward with acceleration $a$, the effective acceleration due to gravity becomes $g_{\text{eff}} = g - a$. Given $L = 1.30 \, \text{m}$, $g = 9.81 \, \text{m/s}^2$, and $a = 1.20 \, \text{m/s}^2$, we find $g_{\text{eff}} = 9.81 - 1.20 = 8.61 \, \text{m/s}^2$. Substituting these values into the formula gives $T = 2\pi\sqrt{\frac{1.30}{8.61}} \approx 2.44 \, \text{s}$.

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