Difference between revisions of "Forest UCM Osc HookesLaw"

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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 equilibrium point is located at the orgin
 
and that  the equilibrium point is located at the orgin
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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> \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>
 
::: <math>= A^{\prime}\cos (\omega t) + B^{\prime} \sin(\omega t)</math>
 
 
let
 
 
:<math>C = \sqrt{\left(A^{\prime}\right)^2+ \left ( B^{\prime}\right)^2}</math>
 
 
:<math>\cos\delta = \frac{A^{\prime}}{C}</math>
 
:<math>\sin\delta = \frac{B^{\prime}}{C}</math>
 
 
then
 
 
:<math>x = A^{\prime}\cos (\omega t) + B^{\prime} \sin(\omega t)</math>
 
::<math> = C \left [ \frac{A^{\prime}}{C}\cos (\omega t) +  \frac{B^{\prime}}{C} \sin(\omega t) \right ]</math>
 
::<math> = C \left [\cos\delta\cos (\omega t) +  \sin\delta \sin(\omega t) \right ]</math>
 
::<math> = C \cos (\omega t - \delta) </math>
 
 
:<math>\cos(A-B) = \cos(A)\cos(B)+\sin(B)\sin(A)</math>
 
 
:<math>x=A \cos(\omega t - \delta)</math>
 
:<math>v = \dot x = \omega A \sin(\omega t - \delta)</math>
 
 
==Kinetic (T) and potential (U) Energy ==
 
 
:<math>U= \frac{1}{2} k x^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>
 
:<math>T= \frac{1}{2} m v^2 =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)</math>
 
 
:<math>U_{\mbox{max}}= \frac{1}{2} k x^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>
 
:<math>T_{\mbox{max}}= \frac{1}{2} m v^2_{\mbox{max}} =\frac{1}{2} m \omega^2 A^2 \sin^2(\omega t - \delta)_{\mbox{max}}= \frac{1}{2} m \omega^2 A^2=U_{\mbox{max}}</math>
 
 
= 2-D Oscillators=
 
 
==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