Forest UCM Osc

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]

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]

Resonance

Forest_Ugrad_ClassicalMechanics