# Kinetic Energy from a Simple Harmonic motion and spring

## 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

BiGyElLoWhAt
Gold Member
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)

cnh1995
Homework Helper
Gold Member
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.

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?

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

Is A= 0.014m?

BiGyElLoWhAt
Gold Member
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.

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 !! :)

BiGyElLoWhAt
Gold Member
No problem.