# Classical mechanics. One-dimensional motion

Here is one task below. How to solve equation $$m\ddot {x} + ax = F(t), x(0) = \dot x (0) = 0$$ in quadratures by using two methods?

I tried to create a system of equations

$$\begin{matrix} \dot v = F(t) - w_{0}^{2}x \\ \dot x = v \\ \end{matrix},$$

but I don't know, what to do next without using some vector $\varepsilon = v + \alpha x$. So, i know only one way. Can you help me with the second way?

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Integrating, i.e. taking the quadrature, twice... I believe you are expected to at least show what you have tried, and where you are stuck, when asking homework questions here. –  Jaime Sep 6 '12 at 20:27
Ok. I solved a system which I wrote in a question by entering some vector $\varepsilon = \alpha x + \beta v$. I can write a solution to you. But I need to solve it without that method. –  PhysiXxx Sep 6 '12 at 20:37
Because it's not an easy method. –  PhysiXxx Sep 6 '12 at 20:58
Your substitution $v = \dot{x}$ is a good starting point, which turns your equation into $m\dot{v}+av = F(t)$. That is a first order linear differential equation, which can be solved easily, see en.wikipedia.org/wiki/…. You will then need to integrate the result once more to get the solution for $x$. –  Jaime Sep 6 '12 at 21:42
Oh, sorry, my equation is $$m\ddot x + ax = F(t).$$ –  PhysiXxx Sep 6 '12 at 21:52

You might not like this answer, but you want to solve for $x\left(t\right)$ in $$m \ddot{x} + \omega^2 x = F\left(t\right),$$ with $\dot{x}\left(0\right) = x\left(0\right) = 0$. A Laplace transform gives you $$x\left(t\right) = \mathcal{L}^{-1}\left\{\frac{\mathcal{L} \left[F\left(t\right)\right] \left(s\right)}{m s^2 + \omega^2}\right\},$$ where we assume $\mathcal{L} \left[F\left(t\right)\right]\left(s\right)$ exists and $F\left(t\right) = 0$ for $t < 0$. Here, $$\mathcal{L} = \int_{0}^{\infty} dt \ e^{-s t},$$ $$\mathcal{L}^{-1} = \frac{1}{2 \pi i}\int_{\mathcal{L}} ds \ e^{s t},$$ and $$\int_{\mathcal{L}} ds$$ denotes integration in the complex $s$ plane along the Laplace contour $\mathcal{L}$. The integral is then evaluated using closure and residue techniques.

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Unfortunately, I don't know math well presently. –  PhysiXxx Sep 7 '12 at 15:12

A lazy way to solve this problem would be to use the same method, but with Matrix calculation...it's basically the same.

Or you can solve $$m\ddot {x} + ax = 0$$

using classical methods, then use the variation of parameters to get the particular solution.

Other solutions would be to use an approximation method, but it could be too complicated.

What type of equation is this ? My guess would be an excited spring or some kind of regulator.

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First solve $$m\ddot {x} + ax = \delta(s)$$ With BC $$x(-\infty) = 0$$ $$\dot x(-\infty) = 0$$
Then integrate this solution to find $$x(t) = \int_0^t F(s)G(s) ds$$