• # question_answer A simple pendulum has a period T on the earth. What is the period T of this same pendulum on the moon, where the acceleration due w gravity is 1/6 that of the earth? A) $\frac{T}{\sqrt{6}}$ B) $\frac{T}{6}$ C) $\sqrt{6}T$ D) 36T

 The length of the pendulum hasn?t changed and the only other factor that determines the period of a pendulum is the acceleration due to gravity $T=2\pi \sqrt{\frac{\ell }{g}}$ $T=2\pi \sqrt{\frac{\ell }{\frac{1}{6g}}}$ $T=\sqrt{6}\left( 2\pi \sqrt{\frac{\ell }{g}} \right)$ $T=\sqrt{6}\,\,T$