What is the equation of motion of a weakly damped oscillator and what is the significance of the results?
For a damped oscillator, we have from Newton's second law:
`(1) m((d^2x)/dt^2) + c(dx/dt) + kx = 0`
Where the forces are
`F_(damp) = -c(dx/dt)` , `F_(elastic) = -kx`
This is just an homogeneous second order differential equation and has solution of the form
`(2)x(t) = e^(lambda t) `
Inserting this solution in (1) we can solve for `lambda` . The following equation is obtained from the substitution:
`mlambda^2 + clambda + k = 0 `
Solving for `lambda` we get
`(3)lambda = (-c+-sqrt(c^2-4mk))/(2m) `
Now, it is clearly that our values for lambda depends on what's inside the square root. Since we wish to solve for a weakly damped oscillator, we have that
`(4)c^2-4mk < 0 `
Because c is very small. This is also known as an under-damped oscillator. Our solution can then be rewritten as (substituting our lambda in (1), using Euler's equation `e^(itheta) = cos(theta) + isen(theta) ` and taking the real part):
`(5)x(t) = A_0e^(-gamma t)cos(wt-alpha) `
With the following
`gamma = c/(2m)` , `w=sqrt(w_0^2 - gamma^2)~~w_0 ` , `w_0 = k/m `
`alpha` and Ao are the phase and amplitude of our oscillation. They are constants determined by the initial conditions.
Looking at the solution for x(t), we can see that the amplitude Ao is modulated by the exponential, which is decreasing in time. So we have an oscillation whose amplitude is decreasing with time. That is, as t goes to infinity, our oscillation goes to zero because
`lim_(t->oo)e^(-gamma t) = 0`
This was expected, because the damping force "removes" energy from our system, thus decreasing the motion of it (and amplitude). I'm attaching a picture of the curve for x(t) to give you a better idea.