The Model[rgb]0,.7,.3Getting Started with Vectors[rgb]0,.7,.3Getting the Forces[rgb]0,.7,.3Putting the Forces Together[rgb]0,.7,.3Finding the Kinetic Energy[rgb]0,.7,.3Finding the Potential Energy[rgb]0,.7,.3Using the Lagrangian[rgb]0,.7,.3Getting Results[rgb]0,.7,.3Getting the Error[rgb]0,.7,.3Types of Motion1/33 Student Projects in Differential Equationshttp://online.redwoods.edu/instruct/darnold/deproj/index.htmA Double Spring PendulumJordan Pierceemail: [email protected]/33 The Modelm1m2(X2, Y2, Z2)(X1, Y1, Z1)Spring 1Spring 23/33 Getting Started with VectorsWhat we know:• The force on mass 1 is always in the direction of the origin andmass 2• The force on mass 2 is always in the direction of mass 1• Gravity always pulls in the -ˆκˆκˆκ direction4/33 What we need:• A unit vector pointing from the origin to mass 1r1= (X1ˆıˆıˆı + Y1ˆˆˆ + Z1ˆκˆκˆκ)kr1k =qX21+ Y21+ Z21ˆr1=r1kr1k=(X1ˆıˆıˆı + Y1ˆˆˆ + Z1ˆκˆκˆκ)pX21+ Y21+ Z215/33 • Unit vectors pointing from mass 2 to mass 1 and mass 1 to mass 2r2= ((X1− X2)ˆıˆıˆı + (Y1− Y2)ˆˆˆ + (Z1− Z2)ˆκˆκˆκ)kr2k =p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2ˆr2=r2kr2k=((X1− X2)ˆıˆıˆı + (Y1− Y2)ˆˆˆ + (Z1− Z2)ˆκˆκˆκ)p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2r3= ((X2− X1)ˆıˆıˆı + (Y2− Y1)ˆˆˆ + (Z2− Z1)ˆκˆκˆκ)kr3k =p(X2− X1)2+ (Y2− Y1)2+ (Z2− Z1)2ˆr3=r3kr3k=((X2− X1)ˆıˆıˆı + (Y2− Y1)ˆˆˆ + (Z2− Z1)ˆκˆκˆκ)p(X2− X1)2+ (Y2− Y1)2+ (Z2− Z1)26/33 Getting the ForcesmxNotice that kr1k is the length of spring 1, stretched or compressed. Theforce of spring 1 on mas s 1 follows the equation F = −kx, where x isthe displacement of the string from un-stretched.F1= −k1(kr1k − L1)r1kr1kF1= k1L1kr1k− 1r17/33 Follow the same steps to get the other forcesF2= −k2(kr2k − L2)r2kr2kF2= k2L2kr2k− 1r2F3= −k2(kr3k − L2)r3kr3kF3= k2L2kr3k− 1r38/33 Putting the Forces TogetherThe force on mass 1 isFm1= F1+ F2− m1gˆκˆκˆκFm1= k1L1kr1k− 1r1+ k2L2kr2k− 1r2− m1gˆκˆκˆκThe force on mass 2 isFm2= F3− m2gˆκˆκˆκFm2= k2L2kr3k− 1r3− m2gˆκˆκˆκ9/33 To find the accelerations of the masses, use the formula F = mam1am1= Fm1am1=Fm1m1m2am2= Fm2am2=Fm2m210/33 So for the acceleration of mass 1 and mass 2am1=k1m1 L1pX21+ Y21+ Z21− 1!(X1ˆıˆıˆı + Y1ˆˆˆ + Z1ˆκˆκˆκ)+k2m1 L2p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2− 1!...∗ ((X1− X2)ˆıˆıˆı + (Y1− Y2)ˆˆˆ + (Z1− Z2)ˆκˆκˆκ) − gˆκˆκˆκam2=k2m2 L2p(X2− X1)2+ (Y2− Y1)2+ (Z2− Z1)2− 1!...∗ ((X2− X1)ˆıˆıˆı + (Y2− Y1)ˆˆˆ + (Z2− Z1)ˆκˆκˆκ) − gˆκˆκˆκ11/33 Let me show theˆıˆıˆı part of am1, which is¨X1=k1m1X1 L1pX21+ Y21+ Z21− 1!+k2m1(X1− X2)...∗ L2p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2− 1!12/33 This is not the only way to get the solution. The Euler Lagrangeequation can also be used. The equation isddt∂`∂ ˙q=∂`∂qWhere ` = T − V , and T = kinetic energy, and V = potential energy.13/33 Finding the Kinetic EnergyKinetic energy = 1/2mv2. For the masses, v = velocity of the massesis just the derivative of their position with respect to time.T =12m1˙X12+12m1˙Y12+12m1˙Z12+12m2˙X22+12m2˙Y22+12m2˙Z22T =12m1˙X12+˙Y12+˙Z12+12m2˙X22+˙Y22+˙Z2214/33 Finding the Potential EnergymxThe p otential energy of a spring is 1/2kx2, where x is the displacementfrom un-stretched.V =12k1qX21+ Y21+ Z21− L12+12k2p(X1− X2)2+ (Y1− Y − 2)2+ (Z1− Z2)2− L22+ m1gZ1+ m2gZ215/33 Using the LagrangianSo the Lagrangian is` =T − V` =12m1˙X12+˙Y12+˙Z12+12m2˙X22+˙Y22+˙Z22−12k1qX21+ Y21+ Z21− L12−12k2p(X1− X2)2+ (Y1− Y − 2)2+ (Z1− Z2)2− L22− m1gZ1− m2gZ216/33 Remember the Euler Lagrange equationddt∂`∂˙q=∂`∂qwhere q is any variable of differentiation. So,∂`∂˙X1=m1˙X1ddt∂`∂˙X1=m1¨X1∂`∂X1=k1X1 L1pX21+ Y21+ Z21− 1!+ k2(X1− X2)...∗ L2p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2− 1!17/33 And doing this for Y1and Z1will yieldm1¨Y1=k1Y1 L1pX21+ Y21+ Z21− 1!+ k2(Y1− Y2)...∗ L2p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2− 1!m1¨Z1=k1Z1 L1pX21+ Y21+ Z21− 1!+ k2(Z1− Z2)...∗ L2p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2− 1!− m1g18/33 Remember that (∂`)/(∂X1) is in theˆıˆıˆı direction. Putting the lastthree equation from the Euler Lagrange equation together, and dividingby m1, I getam1= =k1m1 L1pX21+ Y21+ Z21− 1!(X1ˆıˆıˆı + Y1ˆˆˆ + Z1ˆκˆκˆκ)+k2m1 L2p(X1− X2)2+ (Y1− Y2)2+ (Z1− Z2)2− 1!...∗ ((X1− X2)ˆıˆıˆı + (Y1− Y2)ˆˆˆ + (Z1− Z2)ˆκˆκˆκ) − gˆκˆκˆκWhich is the same as from using vectors to get the acceleration of thefirst mass.19/33 Getting ResultsThese equations can be solved using MATLAB’s ode45 routine. Butthere really is three different models here• When spring 1 has constant zero.• When spring 2 has constant zero.• When both spring constants are non-zero.What will happen when spring 1 has constant zero?20/33 21/33 22/33 Getting the ErrorWe know there was error in this graph because we knew what it shouldlook like. But how do we find the error when we don’t know what itshould look like?• There is no friction• There is no external force• Total energy has to remain constantTotal energy being constant means that the kinetic energy plus thepotential energy has to be constant.E =(TF+ VF) − (TI+ VI)TI+ VI23/33 The error of the first graph is(TF+ VF) − (TI+ VI)TI+ VI= 0.3235The error of the second graph is(TF+ VF) − (TI+ VI)TI+ VI= 0.005724/33 What causes error?−20 −15 −10 −5 0 5 10 15 20−18−16−14−12−10−8−6−4−202X1Z1−20 −15 −10 −5 0 5 10 15 20−20−15−10−505X2Z225/33 0 2 4 6 8 10 12 14 16 18