Previous page : Euler’s step method

Next page : The Earth orbit and Euler’s Step method

Ok, let’s look at

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.

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

so

For *T*=1 s we get

So if we select *k*=10 N/m then

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 EquationsPrevious page : Euler’s step method

Next page : The Earth orbit and Euler’s Step methodLast modified: