A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s²)undergoes simple oscillatory motion.If the spring constant is 12 N/m,find the period of oscillation of this setup on the moon,where g = 1.6 m/s². A)3.35 s B)2.51 s C)4.35 s D)5.70 s

Question

Quizplus · Accepted Answer

The period of oscillation is given by:$$T = 2\pi\sqrt{\frac{m}{k}}$$where m is the mass, k is the spring constant, and g is the acceleration due to gravity.On the Earth, the period of oscillation is:$$T_E = 2\pi\sqrt{\frac{3.42 	ext{ kg}}{12 	ext{ N/m}}}$$$$T_E = 3.35 	ext{ s}$$On the moon, the acceleration due to gravity is 1.6 m/s^2, so the period of oscillation is:$$T_M = 2\pi\sqrt{\frac{3.42 	ext{ kg}}{12 	ext{ N/m}} \frac{9.8 	ext{ m/s}^2}{1.6 	ext{ m/s}^2}}$$$$T_M = 5.70 	ext{ s}$$Therefore, the period of oscillation on the moon is 5.70 s.

A 342 Kg Mass Hanging Vertically from a Spring on the the Earth