Difference between revisions of "Forest UCM Osc HookesLaw"

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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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Returning back to the conservation of energy equation
 
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>
+
:<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)  
 
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> 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> \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
 
:<math> m\ddot x =-kx  </math> energy is constant with time
  
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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
 
 
we define a constant
 
 
:<math> \omega^2  = \frac{k}{m}</math>
 
 
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
 
 
http://www.casaxps.com/help_manual/mathematics/Mechanics3_rev12.pdf
 
 
:<math> \ddot x + \omega^2 x = 0  </math>energy is constant with time
 
 
one could rewrite the above as
 
 
:<math> \left ( \frac{d}{dt}\frac{d}{dt} + \omega^2 \right ) x = 0  </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]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} \gt 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} \; = k [/math]
[math]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]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)


Forest_UCM_Osc#Hooke.27s_Law