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Simple Harmonic Motion-Mass

  • Thread starter physgrl
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  • #1
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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
 

Answers and Replies

  • #2
BruceW
Homework Helper
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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.
 
  • #3
gneill
Mentor
20,792
2,770
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.
 
  • #4
gneill
Mentor
20,792
2,770
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.
 
  • #5
BruceW
Homework Helper
3,611
119
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.

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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