Difference between revisions of "Forest UCM Osc"

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:<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 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).
+
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).
  
 
:<math>\left . \frac{\partial^{2n-1} U}{\partial x^{2n-1}} \right |_{x=x_0} = 0 </math>: no negative inflection
 
:<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>
  
 
==Interpretation (Hooke's law==
 
==Interpretation (Hooke's law==

Revision as of 12:00, 1 October 2014

Hooke's Law

Derivation

In the previous chapter 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 equalibrium 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

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).

[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]

Interpretation (Hooke's law

The 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]


Is this a conservative force?

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)

Potential

[math]U = - \int \vec F \cdot \vec r = - \int (-kx) dx = \frac{1}{2} k x^2[/math]

Simple Harmonic Motion (SHM)

2-D Oscillators

Damped Oscillations

Resonance

Forest_Ugrad_ClassicalMechanics