Deriving Potential Energy and Variance in a Simple Harmonic Oscillator

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Homework Help Overview

The discussion revolves around determining the potential energy of a hanging spring with a mass and spring constant, as well as deriving the mean displacement and estimating the variance of elongation in a simple harmonic oscillator context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expression for potential energy and its derivation, questioning the correctness of their calculations and assumptions regarding mean displacement and elongation.

Discussion Status

Some participants have confirmed the correctness of the potential energy expression, while others are exploring the relationship between force and mean elongation. There is ongoing clarification regarding the interpretation of results and the derivation of variance.

Contextual Notes

Participants are navigating potential sign issues in their equations and the implications of neglecting kinetic energy in their calculations. There is also a focus on the integration limits for estimating variance.

mhellstrom
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Hi all,

I have to determine the potential energy of a hanging spring with a mass m in the end and spring constant k. I try to write down the force in the system

F = m*g + k*x

and integrate the force in order to get the potential energy

E_p = m*g*x+0.5*k*x*x

Does this look correct and is it possible to derive the mean displacement from the potential energy if one could neglect the kinetic energy.

Thanks in advance

Best regards

M
 
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You're expression for E_p is correct. I'm assuming you're taking upward as the positive direction. Your force expression has +/- sign issues, by the way.

The mean displacement is where E_p has a minimum value. So yes, it's possible to derive mean displacement from your E_p expression.
 
Hi,
thanks for the answer. So the mean is when

m*g = k*x

solving for x

x = m*g/k

which results in the mean elongation of the spring is

<dis> = 0.5*m*g/k

Is this correct?

Thanks in advance

all the best
 
mhellstrom said:
Hi,
thanks for the answer. So the mean is when

m*g = k*x

solving for x

x = m*g/k


Correct.

which results in the mean elongation of the spring is

<dis> = 0.5*m*g/k

Is this correct?

Not quite. It contradicts your previous statement.
 
hi,

I am a little bit puzzled where my mistake is... I differentiate my expression for the potential energy in order to find a stationary point

d(E_p) = m*g - k*x

setting this equal to zero and solving for x

x = m*g/k

than I set this into the equation for the potential energy as I presume this is the minimum

E_p = m*g*(m*g/k)-0.5*k*(m*g/k)^2
= 0.5 * (m*g)^2/k

this I would presume is the expression for the mean elongation? Where does I misunderstand thanks in advance

All the best
 
When you get
x = mg/k
you can stop, because that is the mean elongation.
 
thanks,

if I want to estimate the variance of the elongation

var = 1/N sum (xi-x_mean)2

I know the mean is x_mean = m*g/k which I insert into the expression and integrate from minus to plus infinity

var = [tex]\int(m*g-k*x-m*g/k)^2 dx[/tex]

Could anyone give a hint if this is on the right track?

Thanks in advance all the best

M
 

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