Hooke's Law
Derivation
Equation of Motion from Cons of Energy
In the previous chapter Forest_UCM_Energy_Line1D, 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 equilibrium 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
in order to have stable equalibrium
- [math]\left . \frac{\partial^{2} U}{\partial x^{2}} \right |_{x=x_0} \gt 0 [/math]: stable equilibirium
also the odd (2n-1) terms must be zero in order to have stable equilibrium for a symmetric potential ( 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 = T + U = \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
A 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]
While the above was derived from the assumption of conservation of energy we can apply our two tests for conservative forces as a double check:
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)
Equation of motion
In solving the differential equation
- [math] m\ddot x =-kx [/math]energy is constant with time
and observe that the above differential equation is a special case of the more general differential equation
- [math] m\ddot x + c \dot x =-kx [/math]energy is constant with time
one could rewrite the above as
- [math] \left ( \frac{d}{dt}\frac{d}{dt} +\frac{d}{dt} + k \right ) x = 0 [/math]
One could cast the above differential equation into an analogous quadratic equation if you
let
- [math]O = \frac{d}{dt}[/math]
then the analogous equation becomes
- [math] \left ( mO^2 + aO + k \right ) x = 0 [/math]
where m, a, and k are constants
factoring this quadratic you would have
- [math] \left ( O - \gamma \right )\left ( O - \beta \right ) x = 0 [/math]
where a non-trivial solution would exist if one of the terms in the parentheses were zero
this basically reduces our 2nd order differential equation down to two first order differential equations
- [math] \left ( \frac{d}{dt} - \gamma \right ) \left ( \frac{d}{dt} + \beta \right ) x = 0 [/math]
one of the solutions would be
- [math] \frac{d}{dt} x - \gamma x= 0 [/math]
- [math] \frac{dx}{x} = \gamma dt [/math]
- [math] x = Ae^{\gamma t} [/math]
- [math] x = Ae^{\gamma t} + Be^{\beta t} [/math]
For the special case where there isn't a first derivative term (a=0)
You simply have
- [math] \left ( m\frac{d}{dt}\frac{d}{dt} + k \right ) x = 0 [/math]
or
- [math] \left ( mO^2 + k \right ) x = 0 [/math]
- [math] O= \pm \sqrt{-\frac{k}{m}} = \pm i \sqrt{-\frac{k}{m}}[/math]
- [math] \gamma \equiv +i\sqrt{\frac{k}{m}} = i \omega[/math]
- [math] \beta = -i\sqrt{\frac{k}{m}} = -i \omega[/math]
then you have
- [math] x = Ae^{\gamma t} + Be^{\beta t} [/math]
- [math] = Ae^{i \omega t} + Be^{-i\omega t} [/math]
http://www.casaxps.com/help_manual/mathematics/Mechanics3_rev12.pdf
Graph of the Energy
Forest_UCM_Osc#Hooke.27s_Law