Forest UCM Osc

From New IAC Wiki
Jump to navigation Jump to search

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 cas 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 \right |_{x=x)0}[/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