# Kinetic Energy from a Simple Harmonic motion and spring

• Brittany King
Glad I could help.In summary, a 0.26-kg block on a horizontal frictionless surface attached to an ideal massless spring with a spring constant of 190 N/m executes simple harmonic motion when pulled from its equilibrium position at x=0.00m to a displacement of x=+0.080m and released from rest. Using the equations KE=1/2mw^2A^2sin^2(wt+phi), a=(k/m)x, and w=(k/m)^1/2, we can find the kinetic energy of the block at a displacement of x=1.4x10^-2m. To do so, we must first find the amplitude, A, and the phase

## 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!

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)

Brittany King said:
Is there a relationship between the initial equilibrium position for finding A or phi?
You do have A.
Brittany King said:
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
Brittany King said:
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.

BiGyElLoWhAt said:
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
cnh1995 said:
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?

Brittany King said:
To use PE=1/2kx^2 wouldn't I need the x at 0.014m? If that is my A?

Is A= 0.014m?

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.

BiGyElLoWhAt said:
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 ! :)

No problem.