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Simple harmonic oscillator

by mhellstrom
Tags: harmonic, oscillator, simple
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mhellstrom
#1
Oct3-08, 03:18 PM
P: 15
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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Redbelly98
#2
Oct3-08, 05:20 PM
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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.
mhellstrom
#3
Oct4-08, 03:09 PM
P: 15
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

Redbelly98
#4
Oct4-08, 04:34 PM
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P: 12,071
Simple harmonic oscillator

Quote Quote by mhellstrom View Post
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.
mhellstrom
#5
Oct5-08, 02:13 AM
P: 15
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
Redbelly98
#6
Oct5-08, 06:38 AM
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P: 12,071
When you get
x = mg/k
you can stop, because that is the mean elongation.
mhellstrom
#7
Oct5-08, 03:10 PM
P: 15
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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