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

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Homework Help Overview

The discussion revolves around a problem involving a mass undergoing simple harmonic motion attached to a spring. The original poster presents a scenario where a 0.2 kg object is displaced from its equilibrium position, and participants are examining the calculation of its acceleration at a specific displacement.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the impact of gravitational force on the equilibrium position and question the interpretation of displacement from equilibrium. Some suggest using a mathematical approach involving position versus time functions to derive acceleration.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and offering various approaches to calculate acceleration. There is no explicit consensus on the correct interpretation of displacement or the method to use.

Contextual Notes

Some participants note that the equilibrium point shifts due to gravitational forces, which may affect the calculations. There is also mention of the need to clarify the definitions of variables used in the equations.

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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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