What Is the Acceleration of a Mass in Simple Harmonic Motion?

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The discussion centers on calculating the acceleration of a 0.2 kg object in simple harmonic motion, suspended from a spring with a spring constant of 10 N/m, when it is -0.05 m from equilibrium. The initial calculation yields an acceleration of 12.31 m/s², but the correct answer is noted as 2.5 m/s², indicating a misunderstanding of the equilibrium point due to gravitational forces. Participants suggest that the equilibrium point shifts and recommend using sine or cosine functions to derive the position and acceleration expressions. The conversation highlights the importance of correctly identifying displacement from the new equilibrium position in such problems. Understanding these concepts is crucial for accurately solving simple harmonic motion questions.
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Homework Statement



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

Homework Equations



F=ma
F=-kx
F=mg

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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Yeah, this is a tricky question if you don't already know the answer.

The equilibrium point will shift due to the gravity force, so when the object is -0.05m away from equilibrium, this will not be the value of x.

To calculate the new equilibrium point, just solve for a=0. And then you can get the value of x which is -0.05m from the equilibrium point.
 
BruceW said:
Yeah, this is a tricky question if you don't already know the answer.

The equilibrium point will shift due to the gravity force, so when the object is -0.05m away from equilibrium, this will not be the value of x.

Sure it will, if x is the displacement from equilibrium.
 
A straightforward way to approach this problem is to use write an expression for the position versus time (use a sine or cosine function). Find the corresponding acceleration expression (a touch of simple calculus). Take the position expression and rearrange for an expression for t in terms of x. Use it to find acceleration in terms of x.
 
gneill said:
Sure it will, if x is the displacement from equilibrium.
I was referring to her/his? equations, where x is not the displacement from equilibrium. Your post isn't very helpful.

gneill said:
A straightforward way to approach this problem is to use write an expression for the position versus time (use a sine or cosine function). Find the corresponding acceleration expression (a touch of simple calculus). Take the position expression and rearrange for an expression for t in terms of x. Use it to find acceleration in terms of x.
Yep, that's another way of doing it.
 
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