Kinetic Energy from a Simple Harmonic motion and spring

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Homework Help Overview

The problem involves a block attached to a massless spring executing simple harmonic motion. The block is displaced from its equilibrium position and the task is to determine its kinetic energy at a specific displacement.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy in the context of simple harmonic motion. There are attempts to clarify the definitions of amplitude and phase angle, with some questioning how to derive these values from the given conditions.

Discussion Status

Some participants have offered insights into the use of potential energy equations and the implications of initial conditions on kinetic energy. There seems to be an ongoing exploration of the relationships between various parameters, with no explicit consensus reached.

Contextual Notes

There are mentions of specific values for displacement and spring constant, but some participants express uncertainty about the definitions of amplitude and phase angle in this context. The discussion reflects a need for clarity on these concepts without resolving the underlying questions.

Brittany King
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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!
 
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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)
 
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.
 

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