# Homework Help: Damped linear oscillator: Energy losses

1. Mar 14, 2013

### Rulonegger

1. The problem statement, all variables and given/known data
Hello everyone. I need to demonstrate that with a damped free oscillator, which is linear, the total energy is a function of the time, and that the time derivative of the total energy is negative, without saying if the motion is underdamped, critically damped or overdamped, i.e. in general.

2. Relevant equations
From the non-canonical equation for the free oscillator in one dimension $$m\ddot{x}+b\dot{x}+k{x}=0$$ where m is the mass attached to the spring (for example), b is a constant for the damping, and k is the spring constant.

3. The attempt at a solution
I could get the general solution for the equation stated above, but the expressions for the energy are a little cumbersome.
Another way i've been tried is that the equation for the motion of the oscillator times $\dot{x}$ gives
$$m\dot{x}\ddot{x}+b(\dot{x})^2+k{x}\dot{x}=0$$
integrating over time gives
$$\frac{m\ddot{x}^2}{2}+b\int{\dot{x}^2 dt}+\frac{k x^2}{2}=C$$
but the former expression looks like the total energy of the simple harmonic oscillator, which is
$$\frac{m\ddot{x}^2}{2}+\frac{k x^2}{2}=E$$
so i'm tempted to say that the total energy for the damped oscillator is
$$E=C-b\int{\dot{x}^2 dt}$$
and say that if the velocity $\dot{x}$ is not zero ($b>0$ as hypothesis), the expression for E depends explicitly on time, and that the time derivative of E is (using the fundamental theorem for integral calculus):
$$\frac{dE}{dt}=-b\dot{x}^2$$
which is negative, so energy decreases as time increases.
So far, i think everything is correct, but i need to formalize the above steps, first on the fact that $E=C-b\int{\dot{x}^2 dt}$, which was the crucial point of the proof.

2. Mar 14, 2013

### TSny

Everything looks fine to me (except I'm sure you meant to write the kinetic energy as $\frac{m\dot{x}^2}{2}$).

Note that your result for dE/dt makes a lot of sense. The damping force is $f = -b\dot{x}$ and the rate at which that force does work is $f\dot{x} = -b\dot{x}^2$.

3. Mar 15, 2013

### Rulonegger

Thank's TSny, and yeah I've made a copy-paste error. I'm very grateful for your help!