Double Spring Pendulum Jordan Pierce Getting Started Modelling the forces May 10 2004 The Lagrangian Using ode45 Different Models Abstract Home Page I will model the motion of a two spring mass system The first spring is located at the origin with a mass at the other end Off of this mass is another spring with a mass on its end I will be modelling this in 3 D space using vectors and forces and then use the Lagrangian to solve the equations 1 Title Page Getting Started Figure 1 shows the system Spring 1 has spring constant k1 and un stretched length L1 Spring 2 has spring constant k2 and un stretched length L2 Spring 1 is fixed from the origin to m1 and spring 2 is fixed from m1 to m2 In modelling this system I will come up with some impossibilities such as the masses passing through each other or the springs but this can be ignored JJ II J I Page 1 of 16 Go Back The position of the first mass is X1 Y1 Z1 1 X2 Y2 Z2 2 Full Screen And the second mass is located at Close Quit Getting Started Figure 1 The System Modelling the forces The Lagrangian Using ode45 Different Models Home Page Title Page JJ II J I Spring 1 m1 X1 Y1 Z1 Spring 2 Page 2 of 16 Go Back m2 X2 Y2 Z2 Full Screen Close Quit I need to find a few things First the unit vector that points from the first mass to the origin which is the vector that is its position divided by its magnitude r 1 hX1 Y1 Z1 i q krr 1 k X12 Y12 Z12 hX1 Y1 Z1 i r r 1 p 2 X1 Y12 Z12 3 4 5 Getting Started Modelling the forces The Lagrangian This is the direction of the force that spring 1 applies on m1 Spring 2 also applies a force on m1 To find the direction I have to subtract the position of m2 from the position of m1 r 2 hX1 X2 Y1 Y2 Z1 Z2 i p krr 2 k X1 X2 2 Y1 Y2 2 Z1 Z2 2 r r 2 p hX1 X2 Y1 Y2 Z1 Z2 i X1 X2 2 Y1 Y2 2 Z1 Z2 2 r r 3 p hX2 X1 Y2 Y1 Z2 Z1 i X2 X1 2 Y2 Y1 2 Z2 Z1 2 Different Models 6 7 8 Now I must find the direction of the force that spring 2 applies to m2 r 3 hX2 X1 Y2 Y1 Z2 Z1 i p krr 3 k X2 X1 2 Y2 Y1 2 Z2 Z1 2 Using ode45 Home Page Title Page JJ II J I 9 10 11 Now I need to find the forces acting on all of the masses I ll start with m1 there are 4 forces acting on it the force of gravity the force of spring 1 the force of spring 2 and the force of air resistance The force a spring applies is kx where x is the displacement from the resting point of the spring The lengths of spring 1 and spring 2 at rest are L1 and L2 respectively So to find the force of spring 1 on m1 I must find its displacement which is R1 and subtract the springs natural length which is L1 This is x in the formula kx This is the magnitude of the force but to get the force as a vector I must multiply this by r r 1 Page 3 of 16 Go Back Full Screen Close Quit F s1 k1 krr 1 k L1 rr 1 q hX1 Y1 Z1 i X12 Y12 Z12 L1 p 2 F s1 k1 X1 Y12 Z12 L1 1 hX1 Y1 Z1 i F s1 k1 p 2 X1 Y12 Z12 12 13 14 Getting Started Modelling the forces The Lagrangian Now I must model the force spring 2 applies on m1 which will follow the same steps Using ode45 F s2 k2 p p X1 X2 2 Y1 Y2 2 Z1 Z2 2 L2 hX1 X2 Y1 Y2 Z1 Z2 i X1 X2 2 Y1 Y2 2 Z1 Z2 15 Home Page 2 L2 Title Page p 1 X1 X2 2 Y1 Y2 2 Z1 Z2 2 F s2 k2 Different Models 16 JJ II J I hX1 X2 Y1 Y2 Z1 Z2 i And now to find the force spring 2 applies on m2 F s3 k2 F s3 k2 p p Page 4 of 16 X2 X1 2 Y2 Y1 2 Z2 Z1 2 L2 hX2 X1 Y2 Y1 Z2 Z1 i Go Back 17 Full Screen X2 X1 2 Y2 Y1 2 Z2 Z1 2 L2 p 1 X2 X1 2 Y2 Y1 2 Z2 Z1 2 hX2 X1 Y2 Y1 Z2 Z1 i Close 18 Quit Now I must find the force due to air resistance I will assume in this project that the force of air resistance is proportional to the velocity and for sake of experiment I will use two different proportionality constants one for each mass The force from air resistance is always opposite velocity I will now write the vectors in form This will make the analysis easier and gravity is always in the direction The velocities of the masses are 0 0 0 0 0 0 V 1 X1 Y1 Z1 V 2 X2 Y2 Z2 19 Getting Started 20 Modelling the forces The Lagrangian So the force due to air resistance F1r for m1 and F2r for mass two is 0 0 0 0 0 0 F 1R R1 X1 Y1 Z1 F 2R R2 X2 Y2 Z2 2 Using ode45 Different Models 21 22 Modelling the forces Home Page Title Page JJ II J I Figure 2 The force of a spring x m Page 5 of 16 Go Back Now I can model all forces acting on m1 and m2 The force from a spring is kx where x is the displacement from un stretched as seen in figure 2 The force on m1 is Full Screen Close Quit k1 F 1 m1 g k2 L1 p 1 X1 Y1 Z1 X12 Y12 Z12 L2 p 1 X1 X2 2 Y1 Y2 2 Z1 Z2 2 23 Getting Started X1 X2 Y1 Y2 Z1 Z2 0 0 Modelling the forces 0 The Lagrangian R1 X1 Y1 Z1 Using ode45 Different Models Home Page And I can also model all forces acting on m2 F 2 m2 g k2 Title Page L2 p X2 X1 2 Y2 Y1 2 Z2 Z1 2 1 24 X2 X1 Y2 Y1 Z2 Z1 0 0 Please remember that F ma I know the force and I want to find the acceleration The division of m is not shown from this last step to the next but it is trivial Now that I ve got the forces acting on the objects I will separate the forces into the vector components and by dividing by m1 I get the accelerations 00 k1 X1 m1 R1 0 X m1 1 J I Page 6 of 16 …