Forest UCM Osc HookesLaw: Difference between revisions

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Let consider the case where an object is oscillating about a point of stability <math>(x_0)</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
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>
: <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>
Further consider the case the the potential is symmetric about the equilibrium point <math>(x_0)</math>


at the equalibrium point  
at the equilibrium point  


:<math>\left . \frac{\partial U}{\partial x} \right |_{x=x_0} = 0 </math>: Force = 0 at equilibrium
:<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).
in order to have stable equilibrium


:<math>\left . \frac{\partial^{2n-1} U}{\partial x^{2n-1}} \right |_{x=x_0} = 0 </math>: no negative inflection
:<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
and the leading term is just a constant which can be dropped by redefining the zero point of the potential
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if we make the following definitions
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> \left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0} \; = 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>
: <math>U(x) =  \frac{1}{6} \left . \frac{\partial^4 U}{\partial x^4} \right |_{x=x_0} \; = \epsilon </math>


and that  the equailibrium point is located at the orgin
and that  the equilibrium point is located at the orgin


:<math>x_0 = 0</math>  
:<math>x_0 = 0</math>  
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2.) The work done is independent of path ( <math>\vec \nabla \times \vec F = 0</math> in 1-D and 3-D)
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> \gamma = \pm\sqrt{-\frac{k}{m}} \equiv +\sqrt{-\frac{k}{m}} </math>
:<math> \beta = -\sqrt{-\frac{k}{m}} </math>
then you have
:<math> x =  Ae^{\gamma t}  +  Be^{\beta t}  </math>
::<math>  =  Ae^{i \omega t}  +  Be^{-i\omega t}  </math>
:<math> \left ( \frac{d}{dt}\frac{d}{dt} +\frac{d}{dt} + k \right ) x = 0  </math>
http://www.casaxps.com/help_manual/mathematics/Mechanics3_rev12.pdf
:<math> \ddot x + \omega^2 x = 0  </math>energy is constant with time
we define a constant
:<math> \omega^2  = \frac{k}{m}</math>
A solution exists if the term in the parentheses is equal to zero.  An auxilary equation can be defined that is analogous to the derivative operations where
:<math>m \equiv \frac{d}{dt}</math>
:<math> \left ( \frac{d}{dt}\frac{d}{dt}m^2 + \omega^2 \right ) x = 0  </math>
for the term in the parentheses to be zero
:<math> m = \sqrt{- \omega}  </math>
or
:<math> \frac{d}{dt} = \sqrt{- \omega}  </math>
==Graph of the Energy==




[[Forest_UCM_Osc#Hooke.27s_Law]]
[[Forest_UCM_Osc#Hooke.27s_Law]]

Latest revision as of 14:29, 4 October 2021

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

in 1-D

[math]\displaystyle{ \dot {x}^2 = \frac{2}{m} \left ( E-U(x) \right ) }[/math]
[math]\displaystyle{ \dot {x}^2= \frac{2}{m} \left ( E-U(x) \right ) }[/math]
[math]\displaystyle{ \dot {x}= \sqrt{\frac{2}{m} \left ( E-U(x) \right )} }[/math]
[math]\displaystyle{ \frac{dx}{dt}= \sqrt{\frac{2}{m} \left ( E-U(x) \right )} }[/math]
[math]\displaystyle{ \frac{dx}{ \sqrt{\frac{2}{m} \left ( E-U(x) \right )}}=dt }[/math]
[math]\displaystyle{ \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]\displaystyle{ (x_0) }[/math]

A Taylor expansion of the Potential function U(x) about the equilibrium point [math]\displaystyle{ (x_0) }[/math] is

[math]\displaystyle{ 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]\displaystyle{ (x_0) }[/math]

at the equilibrium point

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

in order to have stable equilibrium

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ U(x_0) = 0 }[/math]

This leaves us with

[math]\displaystyle{ 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]\displaystyle{ \left . \frac{\partial^2 U}{\partial x^2} \right |_{x=x_0} \; = k }[/math]
[math]\displaystyle{ U(x) = \frac{1}{6} \left . \frac{\partial^4 U}{\partial x^4} \right |_{x=x_0} \; = \epsilon }[/math]

and that the equilibrium point is located at the orgin

[math]\displaystyle{ x_0 = 0 }[/math]

Then

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

or this 1-D force can be written as

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ E = \frac{m}{2} \dot {x}^2 + \frac{1}{2}kx^2 }[/math]
[math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \vec \nabla \times \vec F = 0 }[/math] in 1-D and 3-D)


Forest_UCM_Osc#Hooke.27s_Law

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