A 3.42-kg stone hanging vertically from an ideal spring on the earth undergoes simple harmonic motion at a place where g = 9.80 m/s². If the force constant (spring constant) of the spring is $12 \mathrm {~N} / \mathrm { m }$ find the period of oscillation of this setup on a planet where g = 1.60 m/s². A) 3.35 s B) 2.51 s C) 4.36 s D) 5.70 s

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Quizplus · Accepted Answer

The period of oscillation of a mass-spring system is given by:$$T = 2\pi \sqrt{\frac{m}{k}}$$where m is the mass of the object, k is the spring constant, and g is the acceleration due to gravity.If the mass and spring constant remain the same, but the acceleration due to gravity changes, then the period of oscillation will change.The new period of oscillation can be found by:$$T' = 2\pi \sqrt{\frac{m}{k}} \sqrt{\frac{g}{g'}}$$where g' is the new acceleration due to gravity.Plugging in the given values, we get:$$T' = 2\pi \sqrt{\frac{3.42 \mathrm{~kg}}{12 \mathrm{~N/m}}} \sqrt{\frac{9.80 \mathrm{~m/s^2}}{1.60 \mathrm{~m/s^2}}} = 3.35 \mathrm{~s}$$Therefore, the correct answer is A.

A 342-Kg Stone Hanging Vertically from an Ideal Spring on the the Earth