2D spring variables

An immoveable (but draggable) anchor point has a spring and bob hanging below and swinging in two dimensions. Regard the bob as a point mass. Define the following variables:

*θ*= angle (0 = vertical, increases counter-clockwise)*S*= spring stretch (displacement from rest length)*L*= length of spring*u*= position of bob*v*=*u*'= velocity of bob*a*=*u*''= acceleration of bob

*R*= rest length of spring*T*= position of anchor point*m*= mass of bob*k*= spring constant*b*= damping constant*g*= gravitational constant

Note that for this simulation the vertical dimension increases downwards.
We'll need the standard unit vectors **i**, **j**. We use bold and overline to indicate a vector.

**i**= unit vector in horizontal direction**j**= unit vector in vertical (down) direction

There are three vector forces acting on the bob:
*m* **a** = **F**_{gravity} + **F**_{spring} + **F**_{damping}
*m* (*a*_{x} **i** + *a*_{y} **j**) = *m* *g* **j** −
*k* *S* (sin *θ* **i** + cos *θ* **j**) −
*b* (*v*_{x} **i** + *v*_{y} **j**)
We can write the horizontal and vertical components of the above as separate equations. This gives us two simultaneous equations. We also divide each side by *m*.

These are the equations of motion. It only remains to show how *S* sin *θ* and *S* cos *θ* are functions of the position of the bob. The displacement of the spring *S* is the current length of the spring minus the rest length.

From the pythagorean theorem we can get the length of the spring *L* in terms of from the position of the bob, *u*, and the position of the anchor point, *T*.

The sine and cosine of the angle *θ* are:

**F**_{gravity}=*m**g***j**= gravity acting straight down**F**_{spring}= −*k**S*(sin*θ***i**+ cos*θ***j**) = the spring pulling (or pushing) along the line from bob to anchor point.**F**_{damping}= −*b*(*v*_{x}**i**+*v*_{y}**j**) = damping (friction) acting opposite to the direction of motion of the bob, ie. opposite to its velocity vector.

a = − _{x}^{k}⁄_{m} S sin θ − ^{b}⁄_{m} v
_{x} | (1a) |

a = _{y}g − ^{k}⁄_{m} S cos θ − ^{b}⁄_{m} v
_{y} | (1b) |

S = L − R
| (2) |

(3) |

sin θ = (u − _{x}T)/_{x}L
| (4a) |

cos θ = (u − _{y}T)/_{y}L
| (4b) |

� Maria Kovalenko, 2007 | PhysLab | Top | Next |