Can a Bead on a Parabolic Wire Exhibit Simple Harmonic Motion?

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Homework Help Overview

The problem involves a bead constrained to move along a parabolic wire described by the equation y = cx². The original poster seeks to demonstrate that the system exhibits simple harmonic motion (SHM) using conservation of energy principles, specifically in the absence of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to express the motion in a specific form related to acceleration and position. The original poster attempts to apply conservation of energy but is uncertain about the correct interpretation of velocity in the context of the bead's motion.

Discussion Status

Some participants have provided insights into the relationship between kinetic and potential energy, while others have clarified the expression for velocity. There is an ongoing exploration of how to derive the necessary equations to show that the system satisfies the conditions for SHM.

Contextual Notes

Participants express confusion regarding the application of derivatives and the relationship between the variables involved. The urgency of the homework deadline is noted, indicating a time constraint affecting the discussion.

Mattofix
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Homework Statement



i have a bead on a wire shaped by y=cx^2 where y is the height - use conservation of energy to show that the system unfergoes simple harmonic motion. - no friction!

Homework Equations



I think to prove its SHM i need to get an equation in the form acc= - w^2 x

The Attempt at a Solution



KE + PE = A constant
1/2mv^2 + mgh = A constant
1/2mv^2 + mgcx^2 = A constant

... is this right? i am pretty sure i need it in the above form them from that i know what w is and can work out the period of this moyion.
 
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please help - with this i can't do the rest of the question :s
 
Well as you said above, you need to show that the system satisfy an equation of the form acc = - constant * position.
More precisly something saying Z'' = - constant * Z. It doesn't matter what this constant exactly is, besides being a constant.
The v in your equation from the the kinetic energy is the SPEED of the bead so it's not the derivative of your x. What is v then? What happens if you plug that expression for v into your equation:

1/2mv^2 + mgcx^2 = constant

then taking the time derivative, so get something with dobbel derivative which is what you are seeking.
 
what do i plug in for v? I am totally confused...
 
It's the length of the velocity vector, so it's
v^2 = x'^2 + y'^2
where x' and y' are the velocities in the x- and y- direction respectivly (ie time derivatives of x and y resp.). What is y' in this case?
 
y' = 2xcx' ?
 
ahhh -= please help somebaody - its in for tomrrow
 
Mattofix said:
y' = 2xcx' ?

This is correct.
 

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