Forest UCM Osc HookesLaw

Hooke's Law

Derivation

Equation of Motion from Cons of Energy

In the previous chapter we saw how the equations of motion could from the requirement that Energy be conserved.

[math]E = T + U[/math]

[math] T = E - U[/math]

[math] \frac{1}{2} m v^2 = E- U[/math]

in 1-D

[math] \dot {x}^2 = \frac{2}{m} \left ( E-U(x) \right )[/math]

[math] \dot {x}^2= \frac{2}{m} \left ( E-U(x) \right )[/math]

[math] \dot {x}= \sqrt{\frac{2}{m} \left ( E-U(x) \right )}[/math]

[math] \frac{dx}{dt}= \sqrt{\frac{2}{m} \left ( E-U(x) \right )}[/math]

[math] \frac{dx}{ \sqrt{\frac{2}{m} \left ( E-U(x) \right )}}=dt[/math]

[math] \sqrt{\frac{m}{2}} \int \frac{dx}{ \sqrt{\left ( E-U(x) \right )}}=\int dt[/math]

Let consider the case where an object is oscillating about a point of stability [math](x_0)[/math]

A Taylor expansion of the Potential function U(x) about the equalibrium point [math](x_0)[/math] is

[math]U(x) = U(x_0) \; + \; \left . \frac{\partial U}{\partial x} \right |_{x=x_0} (x-x_0) \; + \; \frac{1}{2!}\left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0} (x-x_0)^2 \; + \; \frac{1}{3!}\left . \frac{\partial^3 U}{\partial x^3} \right |_{x=x_0} (x-x_0)^3 \; + \dots [/math]

Further consider the case the the potential is symmetric about the equalibrium point [math](x_0)[/math]

at the equalibrium point

[math]\left . \frac{\partial U}{\partial x} \right |_{x=x_0} = 0 [/math]: Force = 0 at equilibrium

also the odd (2n-1) terms must be zero in order to habe stable equalibrium ( if the curvature is negative then the inflection is directed downward towards possibly towards another minima).

[math]\left . \frac{\partial^{2n-1} U}{\partial x^{2n-1}} \right |_{x=x_0} = 0 [/math]: no negative inflection

and the leading term is just a constant which can be dropped by redefining the zero point of the potential

[math]U(x_0) = 0[/math]

This leaves us with

[math]U(x) = \frac{1}{2!}\left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0} (x-x_0)^2 \; + \; \frac{1}{4!}\left . \frac{\partial^4 U}{\partial x^4} \right |_{x=x_0} (x-x_0)^4 \; + \dots [/math]

if we make the following definitions

[math] \left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0} (x-x_0)^2 \; = k [/math]

[math]U(x) = \frac{1}{6} \left . \frac{\partial^4 U}{\partial x^4} \right |_{x=x_0} (x-x_0)^4 \; = \epsilon [/math]

and that the equailibrium point is located at the orgin

[math]x_0 = 0[/math]

Then

[math]U(x) = \frac{1}{2}kx^2 \; + \; \frac{1}{4}\epsilon x^4 \; + \dots [/math]

Since we began this derivation with the assumption that energy was conserved then the force must be conservative such that

[math]: \vec F = - \vec \nabla U[/math]

or this 1-D force can be written as

[math]F = - \frac{\partial }{\partial x} U (x) = - kx - \epsilon x^3 - \dots[/math]

Interpretation (Hooke's law)

Returning back to the conservation of energy equation

[math] E = \frac{m}{2} \dot {x}^2 - \frac{1}{2}kx^2 \; - \; \frac{1}{4}\epsilon x^4 \; + \dots [/math]

Lets consider only the first term in the expansion of the potential U(x)

[math] E = \frac{m}{2} \dot {x}^2 - \frac{1}{2}kx^2 [/math]

[math] \frac{dE{dt} = \frac{m}{2} 2 \dot {x} \ddot x - \frac{1}{2}k2 x \dot x = 0 [/math] energy is constant with time

[math] m\ddot x =-kx [/math] energy is constant with time

The Force exerted by a spring is proportional to the spring displacement from equilibrium and is directed towards restoring the equilibrium condition. (a linear restoring force).

In 1-D this force may be written as

[math]F = - kx[/math]

Is this a conservative force?

1.) The force only depends on position.

2.) The work done is independent of path ( [math]\vec \nabla \times \vec F = 0[/math] in 1-D and 3-D)

Potential

[math]U = - \int \vec F \cdot \vec r = - \int (-kx) dx = \frac{1}{2} k x^2[/math]

Forest_UCM_Osc#Hooke.27s_Law