Deriving Potential Energy and Variance in a Simple Harmonic Oscillator

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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
 
on Phys.org
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
 
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
 
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