## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the period of a 2.0-meter long pendulum on the earth as: $T = 2\pi~\sqrt{\frac{L}{g}}$ $T = 2\pi~\sqrt{\frac{2.0~m}{9.80~m/s^2}}$ $T = 2.84~s$ We can find the length $L_m$ of a pendulum that would have the same period on the moon, which has an acceleration due to gravity of $g_m = 1.63~m/s^2$. Therefore; $T = 2\pi~\sqrt{\frac{L_m}{g_m}}$ $L_m = \frac{T^2~g_m}{(2\pi)^2}$ $L_m = \frac{(2.84~s)^2(1.63~m/s^2)}{(2\pi)^2}$ $L_m = 0.33~m$ The length of the pendulum is 0.33 meters.