View source for Forest UCM Osc HookesLaw

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 equilibrium 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 equilibrium point 

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

in order to have stable equilibrium

:<math>\left . \frac{\partial^{2} U}{\partial x^{2}} \right |_{x=x_0} > 0 </math>: stable equilibirium


also the odd (2n-1) terms must be zero in order to have stable equilibrium for a symmetric potential (the potential needs to be a max at the end points of the motion).

:<math>\left . \frac{\partial^{2n-1} U}{\partial x^{2n-1}} \right |_{x=x_0} = 0 </math>: 

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 equilibrium 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>


==Oscillator properties==

:<math> x =  Ae^{i \omega t}  +  Be^{-i\omega t}   </math>

Using Euler's formula for complex variables

:<math>e^{\pm ix} = \cos(x) \pm  i \sin(x)</math>

:<math>\Rightarrow x = A\left ( \cos(\omega t) + i \sin(\omega t)\right ) + B\left ( \cos(\omega t) - i \sin(\omega t)\right )</math> 
::: <math>= (A + B) \cos (\omega t) + i(A-B) \sin(\omega t)</math>
::: <math>= (A + B) \cos (\omega t) + i(A-B) \sin(\omega t)</math>






http://www.casaxps.com/help_manual/mathematics/Mechanics3_rev12.pdf

==Graph of the Energy==


[[Forest_UCM_Osc#Hooke.27s_Law]]