# Simple harmonic motion again – using Taylor series

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For simple harmonic motion we have that the force is proportional to the displacement from an equilibrium point, and directed in the opposite direction to the displacement. I.e. the object is pulled towards a point, and the pull in proportional to how far away from the point the object is. I.e.

$F = - ky$

Where k is a positive constant and y is the displacement from the equilibrium point. Since we have that F=ma we get

$ma = - ky$

or

$a = - \frac{k}{m}y$

that can be written as

$y'' = - \frac{k}{m}y$

We previously found that the general solution to this is

$y = A\sin \left( {\sqrt {\frac{k}{m}} t - \theta } \right) = A\sin (\omega t - \theta )$

Now let us work with this using Taylor series.

Taylor series

Let us set θ to 0  and let us assume a solution of the form

$y(t) = {a_0} + {a_1}t + {a_2}{t^2} + {a_3}{t^3} + ... = \sum\limits_{k = 0}^\infty {{a_k}{t^k}}$

exists. We then have that

$y'(t) = {a_1} + 2{a_2}t + 3{a_3}{t^2} + ... = \sum\limits_{k = 0}^\infty {k{a_k}{t^{k - 1}}}$

and

$y''(t) = 2{a_2} + 2 \cdot 3{a_3}t + 3 \cdot 4{a_3}{t^2} + ... = \sum\limits_{k = 0}^\infty {k(k - 1){a_k}{t^{k - 2}}}$

So

$2{a_2} + 2 \cdot 3{a_3}t + 3 \cdot 4{a_3}{t^2} + ... = - {\omega ^2}({a_0} + {a_1}t + {a_2}{t^2} + {a_3}{t^3} + ...)$

So

$\begin{gathered} 1 \cdot 2{a_2} = - {\omega ^2}{a_0}, \hfill \\ 2 \cdot 3{a_3} = - {\omega ^2}{a_1}, \hfill \\ 3 \cdot 4{a_3} = - {\omega ^2}{a_2}, \hfill \\ 4 \cdot 5{a_5} = - {\omega ^2}{a_3}, \hfill \\ 5 \cdot 6{a_3} = - {\omega ^2}{a_4}, \hfill \\ ... \hfill \\ \end{gathered}$

That means that if we know a0 and a1 we have the whole series.

So say we have that the solution should go through the origin. We than have a0=0. That means that every second term will be 0. Say we choose a1=1, that gives us

$\begin{gathered} 1 \cdot 2{a_2} = - {\omega ^2}0 \Rightarrow {a_2} = 0 \Rightarrow {a_4} = 0 \Rightarrow ... \hfill \\ 2 \cdot 3{a_3} = - {\omega ^2}1 \Rightarrow {a_3} = \frac{{ - {\omega ^2}}}{{3!}}, \hfill \\ 4 \cdot 5{a_5} = - {\omega ^2}{a_3} = - {\omega ^2}\frac{{ - {\omega ^2}}}{{3!}} \Rightarrow {a_5} = \frac{{{\omega ^4}}}{{5!}}, \hfill \\ ... \hfill \\ \end{gathered}$

So

$\begin{gathered} y(t) = t - \frac{{{\omega ^2}}}{{3!}}{t^3} + \frac{{{\omega ^4}}}{{5!}}{t^5} - ... \hfill \\ \quad \quad = \frac{1}{\omega }\left( {\omega t - \frac{1}{{3!}}{{(\omega t)}^3} + \frac{1}{{5!}}{{(\omega t)}^5} - ...} \right) \hfill \\ \end{gathered}$

So what does the 1/ω stand for?  The derivative of the above stand for the velocity, and for t=0 we get v=ω /ω=1.  We could in principle multiply by an initial velocity v0, and that gives us

$y(t) = \frac{{{v_0}}}{\omega }\left( {\omega t - \frac{1}{{3!}}{{(\omega t)}^3} + \frac{1}{{5!}}{{(\omega t)}^5} - ...} \right)$

In this we can recognize

$y(t) = \frac{{{v_0}}}{\omega }\sin (\omega t) = A\sin (\omega t)$

By starting at a0=A and a0=1 we would end up with the series expansion of

$y(t) = A\cos (\omega t)$

Up a level : Differential Equations
Previous page : Using Taylor series to solve differential equations
Next page : Euler’s step methodLast modified: Mar 30, 2019 @ 12:50