# Damped harmonic motion

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1. Mar 18, 2017

### patrickmoloney

1. The problem statement, all variables and given/known data
Solve the damped harmonic motion system $$\ddot{x} + 2k\dot{x} + \omega^2 x = 0$$
with initial conditions $\dot{x}=V$ at $x = 0$ in the cases
(i)$\, \omega^2 = 10k^2$ (ii) $\omega^2 = k^2$ (iii) $\omega^2 = 5, k = 3$

Identify the type of damping, sketch the curve of $x$ versus $t>0$ in each case, find the zeros of $x(t)$ when applicable
2. Relevant equations

3. The attempt at a solution
$$\ddot{x} + 2k\dot{x} + \omega^2 x = 0$$
To solve this equation we can use the following
$$x = e^{\lambda t}$$ $$\dot{x} = \lambda e^{\lambda t}$$ $$\ddot{x} = \lambda^2 e^{\lambda t}$$
Substitutions these terms into our original equation yields
$$\lambda^2 + 2k\lambda +\omega^2 = 0$$
This is a quadratic equation which can be solved easily

$$\lambda = \dfrac{-2k \pm \sqrt{4k^2 - 4\omega^2}}{2}$$ $$\lambda = \dfrac{-2k \pm 2\sqrt{k^2 - \omega^2}}{2}$$ $$\lambda = -k \pm \sqrt{k^2 - \omega^2}$$
Therefore $\lambda_1 = -k + \sqrt{k^2 - \omega^2}$ and $\lambda_2 = -k - \sqrt{k^2 - \omega^2}$

We have two solutions to the quadratic equation, a linear combination of those solutions will give you the general solution. Therefore, the general solution is $$x(t) = Ae^{\lambda_1 t}+ Be^{\lambda_2 t}$$
$$x(t) = Ae^{(-k + \sqrt{k^2 - \omega^2})t} + Be^{(-k - \sqrt{k^2 - \omega^2})t}$$

(i) $\omega^2 = 10k^2$
In this case $k^2 < \omega^2$ meaning that the term in the square root will result in a negative value(2 imaginary roots).
The two roots are $$\lambda_1 = -k + i\sqrt{\omega^2 -k^2 }$$ $$\lambda_2 = -k - i\sqrt{\omega^2 -k^2}$$
Again using these roots we can get a general solution
$$x(t) = Ae^{(-k + i\sqrt{\omega^2 -k^2})t} + Be^{(-k - i\sqrt{\omega^2 -k^2})t}$$

Using the relationship $$e^{i\theta} = cos\theta + isin\theta$$ and $$\omega' = \sqrt{\omega^2 -k^2}$$
we get $$x(t) = Ae^{-kt}(cos(\omega' t) + \phi)$$

This is an underdamped system since $k^2 < \omega^2$. The curve will oscillate less and less and return to equilibrium. Not too sure about when x(t) will be zero.

(ii) $\omega^2 = k^2$

In this we have a real double root and just let $\omega^2 = k^2$ and substitute that into our general solution. We then use partial fractions to get the solutions $$x(t) = At + Be^{-k}$$ This system is critically damped and the curve looks to decrease exponentially over time until it reaches equilibrium. Again not too sure how to find the zeros.

(iii) $\omega^2 = 5, k = 3$

In this case $k^2 > \omega ^2$ and we substitute in our given values to find out two roots.
$$\lambda _1 = 1$$ $$\lambda_2 = -7$$ Using the values for both of the roots we can find the general solution. $$x(t) = Ae^t + Be^{-7t}$$

The type of damping in this system is over damping. The curve will look like critical damping but it will be wont be as steep and wont reach equilibrium as fast.

So I feel that I'm leaving out quite a bit from the questions here. For part (iii) This is the only case that gives numbers for $k$ and $\omega$, does that mean that this will be the only case where we can find zeros for $x(t)$? For the curves can I just sketch a general curve of what under damping, over damping etc will look like. Or do I actually need to use numerical graphs. Also for part (iii) do I need to find the constants A and B, if so I'm not too sure how to do it a nudge in the right direction would be greatly appreciated. I'm very grateful if you've managed to read this to the end, it's a bit long winded so I tried to skip some algebra parts to save time. Thanks for the help :)

2. Mar 18, 2017

### TJGilb

Are $\omega$ and $k$ not constants? You can always solve for an answer in terms of generic constants. In fact this is more common than being given numbers to work with.

Last edited: Mar 18, 2017
3. Mar 18, 2017

### TJGilb

Alright I've put a bit more effort into looking at this problem now. The first step you can take is determining your relationship between A and B such that you'll only be working with one variable throughout, and once you find one you'll have the other. The very first differential is always true for any time. So, that includes t=0. Start with that.

4. Mar 18, 2017

### TJGilb

So despite my double checking I kept getting answers inconsistent with the original diff eq. So I went back and checked your original process (I had been operating under the assumption you were correct). Everything seems to be mostly good, except I'd go back and check your roots for number three, they're off. When corrected, everything works out and becomes solvable.