MyPhysicsLab – Double Spring 2D

Instructions: click near to square or circles and drag with your mouse

vertical axis:    horizontal axis:    graph type:


An immoveable (but draggable) anchor point has two spring2 and bobs hanging below and swinging in two dimensions. We regard the bobs as point masses. We label the upper spring and bob as number 1, the lower spring and bob as number 2. Define the following variables:

and some constants:

Note that for this simulation the vertical dimension increases downwards.

Here are the equations of motion. The derivation is similar to that given for the single 2D spring.

F1x = m1 a1x = -k1 S1 sin θ1 - b1 v1x + k2 S2 sin θ2
F1y = m1 a1y = -k1 S1 cos θ1 - b1 v1y + k2 S2 cos θ2 + m1 g
F2x = m2 a2x = -k2 S2 sin θ2 - b2 v2x
F2y = m2 a2y = -k2 S2 cos θ2 - b2 v2y + m2 g

The spring stretch Sn and angles θn are functions of the positions un of the bobs as follows:

L1 = √((u1x - Tx)2 + (u1y - Ty)2)
L2 = √((u2x - u1x)2 + (u2y - u1x)2)
S1 = L1 - R1
S2 = L2 - R2
cos θ1 = (u1y - Ty)/L1
sin θ1 = (u1x - Tx)/L1
cos θ2 = (u2y - u1y)/L2
sin θ2 = (u2x - u1x)/L2

Numerical Solution

To solve the equations of motion numerically, so that we can drive the simulation, we use the Runge-Kutta method for solving sets of ordinary differential equations. We need to convert the four second order equations of motion to eight first order equations.

u1x' = v1x
u1y' = v1y
u2x' = v2x
u2y' = v2y
v1x' = -(k1/m1) S1 sin θ1 - (b1/m1) v1x + (k2/m1) S2 sin θ2
v1y' = -(k1/m1) S1 cos θ1 - (b1/m1) v1y + (k2/m1) S2 cos θ2 + g
v2x' = -(k2/m2) S2 sin θ2 - (b2/m2) v2x
v2y' = -(k2/m2) S2 cos θ2 - (b2/m2) v2y + g

We keep in mind that the spring stretch Sn and angles θn are functions of the position of the bob un as given previously.