Finding the Work of a Damped Harmonic Oscillator

[dimensionless]

I'm trying to find the work done by a harmonic oscillator when it moves from $$x_{0} = 0 m$$ to $$x_{max} = 1 m$$.
The oscillator has initial velocity $$v_{0}$$, a maximum height of $$x_{max} = 1 m$$, initial height of $$x_{0} = 0 m$$, a spring constant of $$k$$, a mass of $$m = 1 kg$$, and a damping factor of $$b$$.

It can be represented by the following differential equation:
$$m x'' + b x' + k x = 0$$

Solving for $$x(t)$$ will result in this equation:
$$x = A e^{-\gamma t} sin(\omega_1 t)$$

where $$\gamma = b / (2 m ) = b / 2$$
and $$\omega_1 = ( \omega_0^2 + \gamma^2 )^{1/2}$$
and $$\omega_0 = ( k / m )^{1/2}$$
I think that the potential energy at $$x_{max} = 1 m$$ will equal the initial kinetic energy minus the work required to move the mass to the peak. Work, in general is
$$W = \int F dx$$
I have a spring force $$F_{spring} = - k x$$, but I also have a damping force $$F_{damping} = - b \frac{dx}{dt}$$. I tried working out $$W = \int ( -kx - b\frac{dx}{dt} ) dx$$ but it didn't seem to add up right.

Any ideas on how I can solve $$x_{max} = A e^{-\gamma t} = 1 m$$ in terms of the initial velocity that will be required to get it there?

 

[AlephZero]

I think that the potential energy at $$x_{max} = 1 m$$ will equal the initial kinetic energy minus the work required to move the mass to the peak.

Correct. The easy way to do the question is to use that fact, not to integrate $$W = \int F dx$$

... I have a spring force $$F_{spring} = - k x$$, but I also have a damping force $$F_{damping} = - b \frac{dx}{dt}$$. I tried working out $$W = \int ( -kx - b\frac{dx}{dt} ) dx$$ but it didn't seem to add up right.

The spring force doesn't do any work outside the system, because the point outside the system where the spring is attached doesn't move.

Any ideas on how I can solve $$x_{max} = A e^{-\gamma t} = 1 m$$ in terms of the initial velocity that will be required to get it there?

The same way as you find the max and min of any function - find when the derivative of $$x = A e^{-\gamma t} sin(\omega_1 t)$$
is zero.

NB The way you asked the question seems to assume the maximum occurs when $$sin(\omega_1 t) = 1$$. That assumption is not correct.

 

[dimensionless]

Correct. The easy way to do the question is to use that fact, not to integrate $$W = \int F dx$$



The spring force doesn't do any work outside the system, because the point outside the system where the spring is attached doesn't move.
The same way as you find the max and min of any function - find when the derivative of $$x = A e^{-\gamma t} sin(\omega_1 t)$$
is zero.

Ah! I have an equation for position and another for velocity. That sounds like a much better approach.

NB The way you asked the question seems to assume the maximum occurs when $$sin(\omega_1 t) = 1$$. That assumption is not correct.

It is assumed ...Are saying that the maximum might occur before that?

 

[Hootenanny]

It is assumed ...Are saying that the maximum might occur before that?

Re-read AlephZero's post above. I am assuming you have taken Calc.101 and know that finding the first derivative of a function and setting it to zero gives you the points of the local extrema (i.e min/max). Note, that you should differentiate the whole function as AlephZero suggests, i.e.;

$$\frac{d}{dt} \left[A e^{-\gamma t} sin(\omega_1 t)\right] :=0$$

 

[AlephZero]

If your equations for position and velocity are consistent, then the velocity equation IS the derivative of the position equation

Sure, sometimes people use the approximations that the maximum occurs at time pi/w_1, and w_1 = w_0, and when the damping is small they are pretty good approximations. But you didn't say the damping was small.

Plot exp(-bt)sin(t) for some different values of b (e.g. b = 0, 0.1, 0.5) to see how bad the approximation gets for high levels of damping.