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Kinetic Energy from a Simple Harmonic motion and spring

  • #1

Homework Statement


A 0.26-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 190 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 1.4×10−2 m, what is the kinetic energy of the block?

Homework Equations



KE=1/2mw^2A^2sin^2(wt+phi)
a=(k/m)x
w=(k/m)^1/2


The Attempt at a Solution


I have everything for the equation except for phi and and the amplitude A. I have found a=10.23 m/s2 and w=27.0327rad/s but I am stuck on how to find phi and A. Is there a relationship between the initial equilibrium position for finding A or phi?

Thanks for the help!
 

Answers and Replies

  • #2
BiGyElLoWhAt
Gold Member
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What's wrong with ##PE = 1/2 k x^2##?
You have the amplitude, so you know the total energy. (Yes, you do have the amplitude)
 
  • #3
cnh1995
Homework Helper
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Is there a relationship between the initial equilibrium position for finding A or phi?
You do have A.
The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest
KE=1/2mw^2A^2sin^2(wt+phi
What will be the KE of the block at t=0? Put it in this equation to get phi.
 
  • #4
What's wrong with ##PE = 1/2 k x^2##?
You have the amplitude, so you know the total energy. (Yes, you do have the amplitude)
To use PE=1/2kx^2 wouldn't I need the x at
You do have A.


What will be the KE of the block at t=0? Put it in this equation to get phi.
Is A= 0.014m?
 
  • #5
To use PE=1/2kx^2 wouldn't I need the x at 0.014m? If that is my A?


Is A= 0.014m?
 
  • #6
BiGyElLoWhAt
Gold Member
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No. When it's stretched initially, prior to letting go, what is the KE? At which point in the cycle of a general harmonic oscillator is the KE equal to this value?

Then you're looking for the KE at the point x = .014m. So you know x for the PE, as you know at which point you want to look at.
 
  • #7
No. When it's stretched initially, prior to letting go, what is the KE? At which point in the cycle of a general harmonic oscillator is the KE equal to this value?

Then you're looking for the KE at the point x = .014m. So you know x for the PE, as you know at which point you want to look at.
Oh I got it now!! Thanks so much !! :)
 
  • #8
BiGyElLoWhAt
Gold Member
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113
No problem.
 

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