Simple harmonic motion –Euler’s step method -

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Ok, let’s look at

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

one more time. In this case we need to calculate the acceleration (y´´) given the displacement, then the change of velocity and the velocity (y´) given the acceleration, and finally a new position given the velocity. I.e.

$\begin{gathered} a = - \frac{k}{m}y, \hfill \\ v \leftarrow v + a\Delta t, \hfill \\ y \leftarrow y + v\Delta t \hfill \\ \end{gathered}$

Then the above is repeated over and over again. The arrows stands for “is given by” and is used so that we don’t need to create a new variable name for each new position.

To test this we choose our k and m so that the period will be 1 s. According to the previously found solution we have that

$\omega = \sqrt {\frac{k}{m}} = 2\pi f$

$T = 2\pi \sqrt {\frac{m}{k}}$

For T=1 s we get

${\left( {\frac{1}{{2\pi }}} \right)^2} = \frac{m}{k}$

So if we select k=10 N/m then

$m = 10{\left( {\frac{1}{{2\pi }}} \right)^2} \approx 0.2530{\text{ kg}}$

The below is done in Excel with a time step size of 0.1 s and an initial amplitude of 1 m and using the above mass.

The below is done in Excel with a time step size of 0.1 s and an initial amplitude of 1 m and using the above mass. This is the contents of the first few cells:

	D	E	F	G

3		g	9,81
4		k	10
5		m	0,253302959
6		dt	0,1
7
8	t	y	v	a
9	0	1	0	=-$F$4/$F$5*E9
10	=D9+$F$6	=E9+F10*$F$6	=F9+G9*$F$6	=-$F$4/$F$5*E10

Then row 10 is simply copied down a few hundred cell. This is the result for the first 3 s.

t	y	v	a
0.0	1.0000	0.0000	-39.4784
0.1	0.6052	-3.9478	-23.8930
0.2	-0.0285	-6.3371	1.1251
0.3	-0.6510	-6.2246	25.6989
0.4	-1.0164	-3.6547	40.1273
0.5	-0.9806	0.3580	38.7140
0.6	-0.5577	4.2294	22.0170
0.7	0.0854	6.4311	-3.3719
0.8	0.6948	6.0939	-27.4296
0.9	1.0299	3.3509	-40.6586
1.0	0.9584	-0.7149	-37.8362
1.1	0.5085	-4.4985	-20.0767
1.2	-0.1421	-6.5062	5.6088
1.3	-0.7366	-5.9453	29.0800
1.4	-1.0403	-3.0373	41.0709
1.5	-0.9334	1.0698	36.8476
1.6	-0.4579	4.7545	18.0775
1.7	0.1983	6.5623	-7.8293
1.8	0.7763	5.7793	-30.6453
1.9	1.0477	2.7148	-41.3630
2.0	0.9056	-1.4215	-35.7512
2.1	0.4059	-4.9966	-16.0254
2.2	-0.2540	-6.5991	10.0269
2.3	-0.8136	-5.5964	32.1208
2.4	-1.0521	-2.3844	41.5339
2.5	-0.8752	1.7690	34.5501
2.6	-0.3528	5.2240	13.9264
2.7	0.3089	6.6167	-12.1952
2.8	0.8486	5.3972	-33.5023
2.9	1.0533	2.0469	-41.5832
3.0	0.8422	-2.1114	-33.2478

The Excel file for this can be found here: Simple harmonic motion. It uses decimal comma, so you might have to change that to decimal point in the cells for the constants.

The figure below shows the same in a graphical form. As you can see the period is slightly shorter than 1 s.

With a smaller step size, 0,001 we get the following graph. It is now very close to a sine wave with amplitude 1 and period 1. Matter a fact at 1 s the y value is about 0.99997.

Up a level : Differential Equations
Previous page : Euler’s step method
Next page : The Earth orbit and Euler’s Step method

Last modified: Mar 31, 2019 @ 12:28