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

Student Projects in Differential Equations http://online.redwoods.edu/instruct/darnold/deproj/index.htm 1/33 A Double Spring Pendulum Jordan Pierce � � � � � email: eerrpel@hotmail.com � �

The Model 2/33 � Spring 1 � m 1 ( X 1 , Y 1 , Z 1 ) � Spring 2 � � m 2 ( X 2 , Y 2 , Z 2 ) � �

Getting Started with Vectors 3/33 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 κ ˆ κ direction ˆ � � � � � � �

What we need: • A unit vector pointing from the origin to mass 1 r 1 = ( X 1 ˆ ı ˆ ˆ ı + Y 1 ˆ ı   ˆ ˆ  + Z 1 ˆ κ κ κ ) ˆ ˆ 4/33 � X 2 1 + Y 2 1 + Z 2 � r 1 � = 1 � r 1 � = ( X 1 ˆ r 1 ˆ ˆ ı ı + Y 1 ˆ ı   ˆ ˆ  + Z 1 ˆ κ κ κ ) ˆ ˆ r 1 = ˆ � X 2 1 + Y 2 1 + Z 2 1 � � � � � � �

• Unit vectors pointing from mass 2 to mass 1 and mass 1 to mass 2 r 2 = (( X 1 − X 2 )ˆ ˆ ı ı + ( Y 1 − Y 2 )ˆ ˆ ı    + ( Z 1 − Z 2 )ˆ ˆ ˆ κ κ κ ) ˆ ˆ ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 � � r 2 � = 5/33 � r 2 � = (( X 1 − X 2 )ˆ r 2 ˆ ı ˆ ı ı + ( Y 1 − Y 2 )ˆ   ˆ ˆ  + ( Z 1 − Z 2 )ˆ κ κ ) κ ˆ ˆ r 2 = ˆ ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 � r 3 = (( X 2 − X 1 )ˆ ˆ ı + ( Y 2 − Y 1 )ˆ ı ı ˆ   + ( Z 2 − Z 1 )ˆ  ˆ ˆ κ κ ˆ κ ) ˆ ( X 2 − X 1 ) 2 + ( Y 2 − Y 1 ) 2 + ( Z 2 − Z 1 ) 2 � � r 3 � = � r 3 � = (( X 2 − X 1 )ˆ r 3 ı ˆ ı ˆ ı + ( Y 2 − Y 1 )ˆ   ˆ ˆ  + ( Z 2 − Z 1 )ˆ κ κ κ ) ˆ ˆ � r 3 = ˆ ( X 2 − X 1 ) 2 + ( Y 2 − Y 1 ) 2 + ( Z 2 − Z 1 ) 2 � � � � � � �

Getting the Forces 6/33 x m Notice that � r 1 � 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. � F 1 = − k 1 ( � r 1 � − L 1 ) r 1 � � r 1 � � L 1 � � F 1 = k 1 � r 1 � − 1 r 1 � � � �

Follow the same steps to get the other forces F 2 = − k 2 ( � r 2 � − L 2 ) r 2 � r 2 � � L 2 � 7/33 F 2 = k 2 � r 2 � − 1 r 2 F 3 = − k 2 ( � r 3 � − L 2 ) r 3 � r 3 � � L 2 � F 3 = k 2 � r 3 � − 1 r 3 � � � � � � �

Putting the Forces Together 8/33 The force on mass 1 is F m 1 = F 1 + F 2 − m 1 g ˆ κ κ κ ˆ ˆ � L 1 � L 2 � � F m 1 = k 1 � r 1 � − 1 r 1 + k 2 � r 2 � − 1 r 2 − m 1 g ˆ κ κ κ ˆ ˆ The force on mass 2 is F m 2 = F 3 − m 2 g ˆ κ κ κ ˆ ˆ � � L 2 � � F m 2 = k 2 � r 3 � − 1 r 3 − m 2 g ˆ κ κ ˆ κ ˆ � � � � �

To find the accelerations of the masses, use the formula F = ma m 1 a m 1 = F m 1 a m 1 = F m 1 9/33 m 1 m 2 a m 2 = F m 2 a m 2 = F m 2 m 2 � � � � � � �

So for the acceleration of mass 1 and mass 2 � � a m 1 = k 1 L 1 ˆ ı ˆ  κ ˆ − 1 ( X 1 ˆ ı ˆ ı + Y 1 ˆ   + Z 1 ˆ ˆ κ κ ) ˆ m 1 � X 2 1 + Y 2 1 + Z 2 1 10/33 � � + k 2 L 2 ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 − 1 ... m 1 � ∗ (( X 1 − X 2 )ˆ ı ˆ ˆ ı ı + ( Y 1 − Y 2 )ˆ   ˆ ˆ  + ( Z 1 − Z 2 )ˆ κ κ ) − g ˆ κ ˆ ˆ κ κ ˆ κ ˆ � � a m 2 = k 2 L 2 ( X 2 − X 1 ) 2 + ( Y 2 − Y 1 ) 2 + ( Z 2 − Z 1 ) 2 − 1 ... m 2 � � ∗ (( X 2 − X 1 )ˆ ˆ ı ˆ ı ı + ( Y 2 − Y 1 )ˆ   ˆ ˆ  + ( Z 2 − Z 1 )ˆ κ κ ) − g ˆ κ ˆ ˆ κ κ ˆ κ ˆ � � � � � �

ı Let me show the ˆ ˆ ˆ ı ı part of a m 1 , which is � � X 1 = k 1 L 1 + k 2 ¨ X 1 − 1 ( X 1 − X 2 ) ... m 1 � m 1 X 2 1 + Y 2 1 + Z 2 1 11/33 � � L 2 ∗ ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 − 1 � � � � � � � �

This is not the only way to get the solution. The Euler Lagrange equation can also be used. The equation is d � ∂ℓ � = ∂ℓ dt ∂ ˙ q ∂q 12/33 Where ℓ = T − V , and T = kinetic energy, and V = potential energy. � � � � � � �

Finding the Kinetic Energy 13/33 Kinetic energy = 1 / 2 mv 2 . For the masses, v = velocity of the masses is just the derivative of their position with respect to time. T =1 2 + 1 2 + 1 2 2 m 1 ˙ 2 m 1 ˙ 2 m 1 ˙ X 1 Y 1 Z 1 + 1 2 + 1 2 + 1 2 2 m 2 ˙ 2 m 2 ˙ 2 m 2 ˙ X 2 Y 2 Z 2 T =1 + 1 2 + ˙ 2 + ˙ 2 + ˙ 2 + ˙ � 2 � � 2 � ˙ ˙ 2 m 1 X 1 Y 1 Z 1 2 m 2 X 2 Y 2 Z 2 � � � � � � �

Finding the Potential Energy 14/33 x m The potential energy of a spring is 1 / 2 kx 2 , where x is the displacement from un-stretched. � 2 V =1 �� X 2 1 + Y 2 1 + Z 2 � 2 k 1 1 − L 1 � + 1 � 2 �� ( X 1 − X 2 ) 2 + ( Y 1 − Y − 2) 2 + ( Z 1 − Z 2 ) 2 − L 2 2 k 2 � � + m 1 gZ 1 + m 2 gZ 2 � � �

Using the Lagrangian 15/33 So the Lagrangian is ℓ = T − V ℓ =1 + 1 2 + ˙ 2 + ˙ 2 + ˙ 2 + ˙ � 2 � � 2 � ˙ ˙ 2 m 1 X 1 Y 1 Z 1 2 m 2 X 2 Y 2 Z 2 � 2 �� − 1 2 k 1 X 2 1 + Y 2 1 + Z 2 1 − L 1 − 1 � 2 �� ( X 1 − X 2 ) 2 + ( Y 1 − Y − 2) 2 + ( Z 1 − Z 2 ) 2 − L 2 2 k 2 � � − m 1 gZ 1 − m 2 gZ 2 � � � � �

Remember the Euler Lagrange equation d � ∂ℓ � = ∂ℓ dt ∂ ˙ q ∂q 16/33 where q is any variable of differentiation. So, ∂ℓ = m 1 ˙ X 1 ∂ ˙ X 1 � ∂ℓ � d = m 1 ¨ X 1 ∂ ˙ dt X 1 � � ∂ℓ L 1 � = k 1 X 1 − 1 + k 2 ( X 1 − X 2 ) ... ∂X 1 � X 2 1 + Y 2 1 + Z 2 � 1 � � � L 2 ∗ ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 − 1 � � � � �

And doing this for Y 1 and Z 1 will yield � � L 1 m 1 ¨ Y 1 = k 1 Y 1 − 1 + k 2 ( Y 1 − Y 2 ) ... � X 2 1 + Y 2 1 + Z 2 1 17/33 � � L 2 ∗ ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 − 1 � � � L 1 m 1 ¨ Z 1 = k 1 Z 1 − 1 + k 2 ( Z 1 − Z 2 ) ... � X 2 1 + Y 2 1 + Z 2 1 � � L 2 ∗ ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 − 1 � � � − m 1 g � � � � �

Remember that ( ∂ℓ ) / ( ∂X 1 ) is in the ˆ ˆ ı direction. Putting the last ˆ ı ı three equation from the Euler Lagrange equation together, and dividing by m 1 , I get � � 18/33 a m 1 = = k 1 L 1 ˆ ı  ˆ κ ˆ − 1 ( X 1 ˆ ˆ ı ı + Y 1 ˆ  ˆ  + Z 1 ˆ κ κ ) ˆ � m 1 X 2 1 + Y 2 1 + Z 2 1 � � + k 2 L 2 ( X 1 − X 2 ) 2 + ( Y 1 − Y 2 ) 2 + ( Z 1 − Z 2 ) 2 − 1 ... m 1 � ∗ (( X 1 − X 2 )ˆ ˆ ˆ ı ı + ( Y 1 − Y 2 )ˆ ı   + ( Z 1 − Z 2 )ˆ  ˆ ˆ κ κ ˆ κ ) − g ˆ ˆ κ κ κ ˆ ˆ Which is the same as from using vectors to get the acceleration of the � first mass. � � � � � �

Getting Results 19/33 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? � � � � � � �

20/33 � � � � � � �

21/33 � � � � � � �

Getting the Error 22/33 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. � � � ( T F + V F ) − ( T I + V I ) � � � E = � � T I + V I � � � � �

The error of the first graph is � � ( T F + V F ) − ( T I + V I ) � � � = 0 . 3235 � � T I + V I � 23/33 The error of the second graph is � � ( T F + V F ) − ( T I + V I ) � � � = 0 . 0057 � � T I + V I � � � � � � � �

What causes error? 2 0 −2 −4 24/33 −6 Z 1 −8 −10 −12 −14 −16 −18 −20 −15 −10 −5 0 5 10 15 20 X 1 5 0 −5 � Z 2 −10 � −15 � −20 � −20 −15 −10 −5 0 5 10 15 20 X 2 � � �

50.02 Potential+Kinetic 50.01 25/33 50 49.99 49.98 49.97 49.96 � 49.95 � 0 2 4 6 8 10 12 14 16 18 20 The Potential + Kinetic energy changes only 0.05, or .1 percent � � � � �

50 40 30 20 10 Z 1 0 26/33 −10 −20 −30 −40 −50 −40 −30 −20 −10 0 10 20 30 40 X 1 40 30 20 10 0 Z 2 −10 � −20 −30 � −40 � −50 −50 −40 −30 −20 −10 0 10 20 30 40 X 2 � � � �

4 5.45 x 10 Potential+Kinetic 5.4 27/33 5.35 5.3 5.25 5.2 5.15 � 5.1 0 2 4 6 8 10 12 14 16 18 20 � The Potential + Kinetic energy changes 2500, or 4.5 percent. � � � � �