College of the Redwoods Math 55 Ordinary Differential Equations 1 29 The Motion of Pumping On A Swing Johnathon W Jackson Pumping On A Swing 2 29 Layout Kinetic energy 3 29 Potential energy Lagrangian Euler s formula Approximation of the terms The harmonic system The parametric system Linear and exponential resonance The System 4 29 m3 1 l1 l3 m1 l2 2 m2 Variable Dictionary is the angle in which the system rotates about the origin 5 29 is the angle in which the masses m2 and m3 rotates about the center mass m1 1 is equal to the angle 2 is equal to the sum of the angles Energy K Total kinetic energy 6 29 U Total potential energy L K U Position Coordinates of mass m1 7 29 x1 l1 sin 1 y1 l1 cos 1 Coordinates of mass m2 x2 l1 sin 1 l2 sin 2 y2 l1 cos 1 l2 cos 2 Coordinates of mass m3 x3 l1 sin 1 l3 sin 2 y3 l1 cos 1 l3 cos 2 m3 1 l1 8 29 l3 m1 l2 2 m2 coordinates x1 y1 l1 sin 1 l1 cos 1 x2 y2 l1 sin 1 l2 sin 2 l1 cos 1 l2 cos 2 x3 y3 l1 sin 1 l3 sin 2 l1 cos 1 l3 cos 2 Velocity Velocity of mass m1 9 29 x 1 l1 cos 1 1 y 1 l1 sin 1 1 Velocity of mass m2 x 2 l1 cos 1 1 l2 cos 2 2 y 2 l1 sin 1 1 l2 sin 2 2 Velocity of mass m3 x 3 l1 cos 1 1 l3 cos 2 2 y 3 l1 sin 1 1 l3 sin 2 2 Velocity Squared 2 2 10 29 2 v 1 x 1 y 1 v 22 x 22 y 22 v 32 x 32 y 32 Substituting in the computed values of the x i and y i we have v 2 l2 cos 2 2 l2 sin 2 2 1 1 1 1 1 1 1 v 22 l12 cos 1 2 12 2l1l2 cos 1 cos 2 1 2 l22 cos 2 2 22 l2 sin 1 2 2 2l1l2 sin 1 sin 2 1 2 l2 sin 2 2 2 1 1 2 2 v 32 l12 cos 1 2 12 2l1l3 cos 1 cos 2 1 2 l32 cos 2 2 22 l2 sin 1 2 2 2l1l3 sin 1 sin 2 1 2 l2 sin 2 2 2 1 1 3 2 Kinetic Energy The equation for the kinetic energy is given by the equation 11 29 1 1 1 K m1v12 m2v22 m3v32 2 2 2 Substituting the equations from the previous page in for the squared velocities we get 1 1 K l12 12 m1 m2 m3 22 m2l22 m3l32 2 2 l1 1 2 cos 1 cos 2 sin 1 sin 2 m2l2 m3l3 Potential Energy The equation for the potential energy is given by the equation 12 29 U m1gy1 m2gy2 m1gy3 Substituting the equations for the y coordinates in for y we get U gl1 cos 1 m1 m2 m3 g cos 2 m2l2 m3l3 Substitutions We eliminate a number of parameters for simplicity We let 13 29 M I1 I2 N 1 2 m1 m2 m3 M l12 m2l2 m3l3 m3l3 m2l2 Our Energy Equations Become Kinetic energy 14 29 1 1 K I1 2 I2 2 l1N cos 2 2 Potential energy U M gl1 cos N g cos Lagrangian The Lagrangian for the swing system is 15 29 L K U Substituting the values for the kinetic and potential energies in for K and U we get 1 1 L I1 2 I2 2 l1N cos 2 2 M gl1 cos M N g cos Euler Lagrange Equation The Euler Lagrange Equation is given by the equation d L L 0 dt 16 29 Finally solving for d dt L L we get 0 I1 I2 I2 2l1N cos l1N cos 2l1N sin l1N sin 2 M gl1 sin N g sin cos cos sin Approximations Some of the approximation we have made include the following 17 29 approximation of sin and cosine factors using Taylor series approximation of sin and cosine factors where sin u 0 and cos u 1 for small angles of approximation of sin and cosine factors that are of some degree n to a sum of terms of degree 1 and eliminating terms with frequencies at some multiple of the natural frequency More Substitutions We now make the following substitutions 18 29 1 1 I0 I1 I2 2l1N 1 02 04 4 64 1 1 K0 M gl1 N g 1 02 04 4 64 0 K0 I0 1 1 4 F 0 2I2 N g l1 2 1 02 I0 8 192 0 1 1 A N g 02 04 I0 4 12 1 B l1N 02 04 I0 12 1 1 C l1N 02 04 I0 2 12 The Approximated Swing System The swing system becomes 19 29 u F cos t A cos 2 t B sin 2 t C cos 2 t 1 Notice that the F term isn t dependant on whereas the other three terms on the right hand side are dependant The F term is driving force and the other three terms are parametric terms in that they are functions of the angle they have a time dependant piece The Forced Harmonic System We can approximate the motion of the system when is small by looking at just F term Thus our new system becomes 20 29 u F cos t When solving the differential equation for with initial conditions that the swing initially at rest at time t 0 0 0 we get u F t 2 0 sin 0t 21 29 which results in a linear growth per cycle D F 02 Graph of phi vrs t 300 22 29 200 phi 100 0 100 200 300 0 5 10 15 20 25 t 30 35 40 45 50 Graph of phi vrs t 400 23 29 300 200 phi 100 0 100 200 300 0 5 10 15 20 25 t 30 35 40 45 50 Solving For The Parametric System We can approximate the motion of the system when is large by looking at the dependant terms Thus our new system becomes 24 29 u A cos 2 t B sin 2 t C cos 2 t The difference between the forced harmonic oscillator system and the dependant system is that an initial condition at time t 0 results in no amplitude growth in the system Thus we have to assume that the swing is pulled back to some initial angle of for there to be some positive contribution to the growth of the amplitude of the system When solving the differential equation for we get 2 u 2 a e A 0B 0 C t 4 0 cos t sin t 25 29 which results in an exponential growth per cycle P 2 0 0 Graph of phi vrs t 2000 1500 26 29 1000 phi 500 0 500 1000 1500 2000 …