A simple pendulum has a period of 3.65 s on the surface of the Earth.If this pendulum were placed on the surface of the Moon,where the gravitational acceleration is 1.62 m/s²,what would its period be? A)1.48 s B)8.98 s C)16.4 s D)22.1 s E)33.3 s

Question

Quizplus · Accepted Answer

The period of a pendulum is given by T=2π√(L/g), where L is the length of the pendulum and g is the gravitational acceleration. Since the length of the pendulum is not given and does not affect the ratio of periods between Earth and the Moon, we can use the ratio T_Moon/T_Earth=√(g_Earth/g_Moon). Plugging in the values given, we get T_Moon/T_Earth=√(9.81 m/s^2/1.62 m/s^2)=2.47. Thus, the period on the Moon would be T_Moon=2.47 T_Earth=2.47(3.65 s)=9.02 s, which is closest to choice B.

A Simple Pendulum Has a Period of 3