Deriving Potential Energy and Variance in a Simple Harmonic Oscillator

In summary: The expression for the variance of the elongation is var = 1/N sum (xi-x_mean)2. So, to estimate the variance of the elongation, you would integrate the expression from minus to plus infinity.
  • #1
mhellstrom
15
0
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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
When you get
x = mg/k
you can stop, because that is the mean elongation.
 
  • #7
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
 

What is a simple harmonic oscillator?

A simple harmonic oscillator is a physical system that exhibits a repetitive or oscillatory motion around an equilibrium point. Examples include a mass on a spring or a pendulum.

What is the equation of motion for a simple harmonic oscillator?

The equation of motion for a simple harmonic oscillator is given by x(t) = A cos(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

What is the relationship between the period and frequency of a simple harmonic oscillator?

The period (T) of a simple harmonic oscillator is the time it takes to complete one full cycle of oscillation, while the frequency (f) is the number of cycles per unit time. They are related by the equation T = 1/f.

What factors affect the period and frequency of a simple harmonic oscillator?

The period and frequency of a simple harmonic oscillator are affected by the mass of the object, the stiffness of the spring or restoring force, and the initial conditions (amplitude and phase angle).

What is the relationship between potential energy and kinetic energy in a simple harmonic oscillator?

In a simple harmonic oscillator, the potential energy is maximum at the equilibrium point and decreases as the object moves away from equilibrium. At the same time, the kinetic energy is minimum at the equilibrium point and increases as the object moves away from equilibrium. At any point in the oscillation, the sum of the potential and kinetic energy is constant and equal to the total mechanical energy of the system.

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