(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 0.2 kg object is suspended from a spring with a spring constant of k=10 N/m and is undergoing simple harmonic motion. What is its acceleration of the object at the instant when it is -0.05 m away from equilibrium?

A. 1000 m/s2

B. 40 m/s2

C. 0.1 m/s2

D. 2.5 m/s2

2. Relevant equations

F=ma

F=-kx

F=mg

3. The attempt at a solution

F=F(gravity)+F(spring)

ma=mg+(-kx)

0.20kg*a=(0.20kg*9.81m/s^2)+(-(10N/m*-0.05m))

a=12.31m/s^2

The answer is supposed to be D

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# Homework Help: Simple Harmonic Motion-Mass

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