[PPT] - A Double Spring Pendulum Jordan Pierce email: PowerPoint Presentation

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Student Projects in Differential Equations

http://online.redwoods.edu/instruct/darnold/deproj/index.htm

A Double Spring Pendulum

Jordan Pierce

email: eerrpel@hotmail.com

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The Model

m1 m2 (X2, Y2, Z2) (X1, Y1, Z1) Spring 1 Spring 2

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Getting Started with Vectors

What we know:

The force on mass 1 is always in the direction of the origin and

mass 2

The force on mass 2 is always in the direction of mass 1
Gravity always pulls in the -ˆ

κ ˆ κ ˆ κ direction

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What we need:
A unit vector pointing from the origin to mass 1

r1 = (X1ˆ ı ˆ ı ˆ ı + Y1ˆ  ˆ  ˆ  + Z1ˆ κ ˆ κ ˆ κ) r1 =

X2

1 + Y 2 1 + Z2 1

ˆ r1 = r1 r1 = (X1ˆ ı ˆ ı ˆ ı + Y1ˆ  ˆ  ˆ  + Z1ˆ κ ˆ κ ˆ κ)

X2

1 + Y 2 1 + Z2 1

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Unit vectors pointing from mass 2 to mass 1 and mass 1 to mass 2

r2 = ((X1 − X2)ˆ ı ˆ ı ˆ ı + (Y1 − Y2)ˆ  ˆ  ˆ  + (Z1 − Z2)ˆ κ ˆ κ ˆ κ) r2 =

(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2

ˆ r2 = r2 r2 = ((X1 − X2)ˆ ı ˆ ı ˆ ı + (Y1 − Y2)ˆ  ˆ  ˆ  + (Z1 − Z2)ˆ κ ˆ κ ˆ κ)

(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2

r3 = ((X2 − X1)ˆ ı ˆ ı ˆ ı + (Y2 − Y1)ˆ  ˆ  ˆ  + (Z2 − Z1)ˆ κ ˆ κ ˆ κ) r3 =

(X2 − X1)2 + (Y2 − Y1)2 + (Z2 − Z1)2

ˆ r3 = r3 r3 = ((X2 − X1)ˆ ı ˆ ı ˆ ı + (Y2 − Y1)ˆ  ˆ  ˆ  + (Z2 − Z1)ˆ κ ˆ κ ˆ κ)

(X2 − X1)2 + (Y2 − Y1)2 + (Z2 − Z1)2

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Getting the Forces

m x

Notice that r1 is the length of spring 1, stretched or compressed. The force of spring 1 on mass 1 follows the equation F = −kx, where x is the displacement of the string from un-stretched. F1 = −k1(r1 − L1) r1 r1 F1 = k1 L1 r1 − 1

r1

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Follow the same steps to get the other forces

F2 = −k2(r2 − L2) r2 r2 F2 = k2 L2 r2 − 1

r2

F3 = −k2(r3 − L2) r3 r3 F3 = k2 L2 r3 − 1

r3

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Putting the Forces Together

The force on mass 1 is Fm1 = F1 + F2 − m1gˆ κ ˆ κ ˆ κ Fm1 = k1 L1 r1 − 1

r1 + k2

L2 r2 − 1

r2 − m1gˆ

κ ˆ κ ˆ κ The force on mass 2 is Fm2 = F3 − m2gˆ κ ˆ κ ˆ κ Fm2 = k2 L2 r3 − 1

r3 − m2gˆ

κ ˆ κ ˆ κ

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To find the accelerations of the masses, use the formula F = ma

m1am1 = Fm1 am1 = Fm1 m1 m2am2 = Fm2 am2 = Fm2 m2

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So for the acceleration of mass 1 and mass 2

am1 = k1 m1

L1
X2

1 + Y 2 1 + Z2 1

− 1

(X1ˆ

ı ˆ ı ˆ ı + Y1ˆ  ˆ  ˆ  + Z1ˆ κ ˆ κ ˆ κ) + k2 m1

L2
(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2 − 1
...

∗ ((X1 − X2)ˆ ı ˆ ı ˆ ı + (Y1 − Y2)ˆ  ˆ  ˆ  + (Z1 − Z2)ˆ κ ˆ κ ˆ κ) − gˆ κ ˆ κ ˆ κ am2 = k2 m2

L2
(X2 − X1)2 + (Y2 − Y1)2 + (Z2 − Z1)2 − 1
...

∗ ((X2 − X1)ˆ ı ˆ ı ˆ ı + (Y2 − Y1)ˆ  ˆ  ˆ  + (Z2 − Z1)ˆ κ ˆ κ ˆ κ) − gˆ κ ˆ κ ˆ κ

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Let me show the ˆ

ı ˆ ı ˆ ı part of am1, which is ¨ X1 = k1 m1 X1

L1
X2

1 + Y 2 1 + Z2 1

− 1

+ k2

m1 (X1 − X2)... ∗

L2
(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2 − 1

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This is not the only way to get the solution. The Euler Lagrange

equation can also be used. The equation is d dt ∂ℓ ∂ ˙ q

= ∂ℓ

∂q Where ℓ = T − V , and T = kinetic energy, and V = potential energy.

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Finding the Kinetic Energy

Finding the Potential Energy

m x

The potential energy of a spring is 1/2kx2, where x is the displacement from un-stretched. V =1 2k1

X2

1 + Y 2 1 + Z2 1 − L1

2 + 1 2k2

(X1 − X2)2 + (Y1 − Y − 2)2 + (Z1 − Z2)2 − L2

2 + m1gZ1 + m2gZ2

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Using the Lagrangian

So the Lagrangian is ℓ =T − V ℓ =1 2m1

˙

1 + Y 2 1 + Z2 1 − L1

2 − 1 2k2

(X1 − X2)2 + (Y1 − Y − 2)2 + (Z1 − Z2)2 − L2

2 − m1gZ1 − m2gZ2

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Remember the Euler Lagrange equation

d dt ∂ℓ ∂ ˙ q

= ∂ℓ

∂q where q is any variable of differentiation. So, ∂ℓ ∂ ˙ X1 =m1 ˙ X1 d dt ∂ℓ ∂ ˙ X1

=m1 ¨

X1 ∂ℓ ∂X1 =k1X1

L1
X2

1 + Y 2 1 + Z2 1

− 1

+ k2(X1 − X2)...

∗

L2
(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2 − 1

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And doing this for Y1 and Z1 will yield

m1 ¨ Y1 =k1Y1

L1
X2

1 + Y 2 1 + Z2 1

− 1

+ k2(Y1 − Y2)...

∗

L2
(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2 − 1
m1 ¨

Z1 =k1Z1

L1
X2

1 + Y 2 1 + Z2 1

− 1

+ k2(Z1 − Z2)...

∗

L2
(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2 − 1
− m1g

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Remember that (∂ℓ)/(∂X1) is in the ˆ

ı ˆ ı ˆ ı direction. Putting the last three equation from the Euler Lagrange equation together, and dividing by m1, I get am1 = = k1 m1

L1
X2

1 + Y 2 1 + Z2 1

− 1

(X1ˆ

ı ˆ ı ˆ ı + Y1ˆ  ˆ  ˆ  + Z1ˆ κ ˆ κ ˆ κ) + k2 m1

L2
(X1 − X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2 − 1
...

∗ ((X1 − X2)ˆ ı ˆ ı ˆ ı + (Y1 − Y2)ˆ  ˆ  ˆ  + (Z1 − Z2)ˆ κ ˆ κ ˆ κ) − gˆ κ ˆ κ ˆ κ Which is the same as from using vectors to get the acceleration of the first mass.

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Getting Results

These equations can be solved using MATLAB’s ode45 routine. But there 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?

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Getting the Error

We know there was error in this graph because we knew what it should look like. But how do we find the error when we don’t know what it should look like?

There is no friction
There is no external force
Total energy has to remain constant

Total energy being constant means that the kinetic energy plus the potential energy has to be constant. E =

(TF + VF) − (TI + VI)

TI + VI

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The error of the first graph is
(TF + VF) − (TI + VI)

TI + VI

= 0.3235

The error of the second graph is

(TF + VF) − (TI + VI)

TI + VI

= 0.0057

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What causes error?

−20 −15 −10 −5 5 10 15 20 −18 −16 −14 −12 −10 −8 −6 −4 −2 2 X1 Z1 −20 −15 −10 −5 5 10 15 20 −20 −15 −10 −5 5 X2 Z2

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2

4 6 8 10 12 14 16 18 20 49.95 49.96 49.97 49.98 49.99 50 50.01 50.02

Potential+Kinetic

The Potential + Kinetic energy changes only 0.05, or .1 percent

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−40

−30 −20 −10 10 20 30 40 −50 −40 −30 −20 −10 10 20 30 40 50 X1 Z1 −50 −40 −30 −20 −10 10 20 30 40 −50 −40 −30 −20 −10 10 20 30 40 X2 Z2

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2

4 6 8 10 12 14 16 18 20 5.1 5.15 5.2 5.25 5.3 5.35 5.4 5.45 x 10

4

Potential+Kinetic

The Potential + Kinetic energy changes 2500, or 4.5 percent.

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−20

−15 −10 −5 5 10 15 20 −18 −16 −14 −12 −10 −8 −6 −4 −2 X1 Z1 −20 −15 −10 −5 5 10 15 20 −20 −15 −10 −5 5 X2 Z2

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2

4 6 8 10 12 14 16 18 20 −0.1 0.1 0.2 0.3 0.4 0.5 0.6

Potential+Kinetic

The Potential + Kinetic energy changes 0.55, or ∞ percent.

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2

4 6 8 10 12 14 16 18 20 −800 −600 −400 −200 200 400 600 800 Potential Kinetic

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−25

−20 −15 −10 −5 5 10 15 20 25 −25 −20 −15 −10 −5 5 X1 Z1

Figure 1: Periodic Motion

Types of Motion

As seen from before, with just the two masses and one spring, the motion is circular. The motion of that system will always be circular. When spring 2 has constant zero, the motion will be periodic.

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With both spring constants non-zero, the motion can be quite chaotic,

so I made an animating program to view it easier. The amount that it is chaotic is dependent on how much the masses affect each other. If mass 1 is large, and mass 2 is small, the motion of mass 1 will be more periodic.