Coupled Masses and Modes of Oscillation

Todd Crutcher

June 10, 2003

Abstract

This research paper will investigate the different stable modes of oscillation for coupled masses. In particular Iwill concentrate on two and three pendulum systems that are coupled at the bottom by springs. I will give examplesof both analytical and experimental results for these systems, as well as attempting to derive a general formula for thenumber of modes.

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1 Introduction

In order to study the modes of oscillation for coupled mass pendulums it was first necessary to construct a set ofcoupled mass pendulums. For this I built both a two, and a three Pendulum system. While observing the apparentmodes of operation of the two and three pendulum systems (3 and 6 respectively) it appeared that number of stablemodes for a system followed the pattern:

1�

2�����������

n � 1 � � n (1)

It will later be shown from the theoretical and analytical results that this is incorrect. The actual number of stablemodes for n pendulums is:

n! (2)

Results for these experiments were obtained both theoretically and experimentally.

2 The Experiment

I constructed, using the lab pro equipment and the logger pro software, both a two and three pendulum system thatallowed for recording position data directly. This was done by using the rotary motion sensors as the pivot points forthe pendulums.

Figure 1: picture of the three Pendulum experiment

The lab pro hardware is capable of connecting two rotary motion sensors at once. So for the three pendulum systemI was forced to use two computers and four sensors. I then coupled two sensors together and sent there outputs toseparate computers. These were later used as index points to combine all the data in a spreadsheet.

3 Theory

In order to theoretically study coupled mass pendulums I first had derive the equations of motion for the particularsystem that we are interested in. To do this I had to make the following assumptions.

The pendulums all are the same mass.

The springs have no mass and equal constants ”k”.

The springs are attached to the center of the mass.

The arm of the pendulum has no mass.

Because of the large size of weight that is used in the experiment (2Kg) it is safe to ignore the mass of the springs andconnecting arms. Next using polar coordinates and the sum of the forces is equals the mass times acceleration in eachdirection.

2

∑Fθ maθ (3)

∑Fr mar (4)

3.1 2-Pendulums

θ1

θ1

θ 2

θ 2

θ1

θ 2

mg sin mg sin

−ks cos

ks cos

Figure 2: Two Pendulum Free Body Diagram

I was able to obtain the following sets of differential equations for the motion of the coupled mass pendulum systems.

KS�cos

� θ1 ��� � mg�sin

� θ1 ��� mrθ̈1 (5)

� KS�cos

� θ2 ��� � mg�sin

� θ2 ��� mrθ̈2 (6)

Where K is the spring constant. S is the amount that the spring is stretched.

S D�

L�sin

� θ2 ��� sin� θ1 ��� (7)

D is the distance between the pivot points minus the un-stretched length of the spring. these equations are easily solvedusing the Runge-Kutta method and a c++ program, thus obtaining a numerical solution to this system (see results).

3.2 3-Pendulums

θ1

θ1

θ 2

θ 2

mg sin

−ks cos θ 2

θ1ks cos

3θ

3θ

3θ

mg sin mg sin

1 ks cos θ

1

22

2−ks cos

Figure 3: Three Pendulum Free Body Diagram

It is now only a minor task to expand the previous set of equations to include a third pendulum. this was done by onceagain starting with a drawing of the system and thus obtaining the following set of differential equations.

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KS1�cos

� θ1 ��� � mg�sin

� θ1 ��� mrθ̈1 (8)

K�S2 � S1 � cos

� θ2 � � mgsin� θ2 � mrθ̈2 (9)

� KS2�cos

� θ3 ��� � mg�sin

� θ3 ��� mrθ̈3 (10)

Where K is again the spring constant. S1 and S2 are the amount that the spring is stretched.

S1 D�

L�sin

� θ2 � � sin� θ1 ��� (11)

S2 D�

L�sin

� θ3 � � sin� θ2 ��� (12)

D is still the distance between the pivot points minus the un-stretched length of the spring. These equations were againsolved numerically using the Runge-Kutta method and a c++ program (see Results)

3.3 friction

θ1

θ1

θ 2

θ 2

mg sin

−ks cos θ 2

θ1ks cos

3θ

3θ

3θ

F1F2 F3

mg sin mg sin

1 ks cos θ

1

22

2−ks cos

Figure 4: Three Pendulum with Friction Free Body Diagram

Next I went back to the free body diagram’s and added a frictional force opposing the motion of the pendulum. Fromthere I then reworked the previous set of equations for the three pendulum system. This yielded the following set ofdifferential equations.

KS1�cos

� θ1 ��� � F1�

mg�sin

� θ1 ��� mrθ̈1 (13)

K�S2 � S1 � cos

� θ2 � � F2�

mgsin� θ2 � mrθ̈2 (14)

� KS2�cos

� θ3 ��� � F3�

mg�sin

� θ3 ��� mrθ̈1 (15)

Where S1 and S2 are the same as before and F1 , F2, and F3 are as follows.

F1 KS1�sin

� θ1 ��� � mg�cos

� θ1 ��� µk (16)

F2 K�S2 � S1 � sin

� θ2 ����� mg�cos

� θ2 ��� µk (17)

F3 � KS2�sin

� θ3 ��� � mg�cos

� θ3 ��� µk (18)

4 The Program

The c++ program that was used is a standard fourth-order Runge-Kutta routine. The program was set to take commandline inputs to determine the initial conditions. The mass of the weight on the end of the pendulum is 2Kg and is muchgrater than the mass of the springs and connecting arms. It is then safe to only consider the mass of the pendulumsweight and ignore the mass of the springs and connecting arms.

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5 Results

5.1 2-pendulum

The following three figures show the motion of the two pendulums (no friction) on the right. On the left is a drawingof the initial position of the system. These were initial believed to be the three stable modes of oscillation for thissystem. However after closer inspection the figures ?? and ?? are actually the same mode, it is only the magnitude ofthe motion that is different.

0 20 40 60 80 100Time(s)

Run 1

Figure 5: graphical representation of the first set of initial conditions

0 20 40 60 80 100Time (s)

Run 2

Figure 6: graphical representation of the second set of initial conditions

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0 20 40 60 80 100Time (s)

Run 3

Figure 7: graphical representation of the third set of initial conditions

5.2 3-pendulum

The following six figures show the motion of the three pendulums (with friction included) on the right. On the left is adrawing of the initial position of the system. These represent the six stable modes of oscillation for the six pendulumsystem.

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0 20 40 60 80 100 120Time (s)

Run 1

Figure 8: graphical representation of the first set of initial conditions

0 20 40 60 80 100 120Time (s)

Run 2

Figure 9: graphical representation of the second set of initial conditions

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0 20 40 60 80 100Time (s)

Run 3

Figure 10: graphical representation of the third set of initial conditions

0 20 40 60 80 100Time (s)

Run 4

Figure 11: graphical representation of the fourth set of initial conditions

8

0 20 40 60 80 100 120Time (s)

Run 5

Figure 12: graphical representation of the fifth set of initial conditions

0 20 40 60 80 100 120Time (s)

Run 6

Figure 13: graphical representation of the sixth set of initial conditions

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6 Conclusion

After realizing that the 2 pendulum system has only two stable modes, and three pendulums have six stable modes, itappears that the number of stable modes for n pendulums follows n!. Further work will need to be done however toprove this.

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7 Appendix A

The following code is the C++ programs used for this project.

7.1 2 pendulums

#include <iostream.h>#include <stdio.h>#include <math.h>

#define N 4#define DELTA_T 0.05#define T_MAX 100.0

#define D .25#define K 3.5#define m 2#define g 9.81#define L .43

int main(int argc, char **argv){

double t, y[20],inputs[argc];int i,j,k;float phi_1,phi_2;void runge4(double x, double y[], double step); // Runge-Kutta function

if (argc<=2){

cout << "\n\n\nYou need to include command line arguments!!\n" << endl;cout << "\nExample: ./a.out <phi_1> <phi_2>!!\n\n\n" << endl;return 0;

}else{

sscanf(argv[1], "%f", &phi_1);sscanf(argv[2], "%f", &phi_2);

}

y[0]= phi_1;y[1]= 0.0;y[2]= phi_2;y[3]= 0.0;

for (j=1; j*DELTA_T <= T_MAX ;++j) {t=j*DELTA_T;runge4(t, y, DELTA_T);

printf("%lf\t", j*DELTA_T); //printf("%lf\t", y[0]+1); //printf("%lf\t", y[2]-1); //printf("\n");

}

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return 0;}//end of main

double f(double x, double y[], int i) {

if (i==0) return(y[1]);

if (i==1) return((( (K*D)-(K*L*sin(y[0])) + (K*L*sin(y[2])) )*cos(y[0])+(m*g*sin(y[0])))/-L);

if (i==2) return(y[3]);

if (i==3) return((-( (K*D)-(K*L*sin(y[0])) + (K*L*sin(y[2])) )*cos(y[2])+(m*g*sin(y[2])))/-L);

}//end of f

void runge4(double x, double y[], double step) {double h=step/2.0;double t1[N], t2[N], t3[N];double k1[N], k2[N], k3[N],k4[N];int i;

for (i=0;i<N;i++) t1[i]=y[i]+0.5*(k1[i]=step*f(x, y, i));for (i=0;i<N;i++) t2[i]=y[i]+0.5*(k2[i]=step*f(x+h, t1, i));for (i=0;i<N;i++) t3[i]=y[i]+(k3[i]=step*f(x+h, t2, i));for (i=0;i<N;i++) k4[i]=step*f(x+step, t3, i);

for (i=0;i<N;i++) y[i]+=(k1[i]+2*k2[i]+2*k3[i]+k4[i])/6.0;return;

}//end of runge4

7.2 3 pendulums

#include <iostream.h>#include <stdio.h>#include <math.h>

#define N 6#define DELTA_T 0.005#define T_MAX 200.0

#define D .25#define K 3.5#define m 2#define g -9.81#define L .4#define r .43

static float u;int main(int argc, char **argv){

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double t, y[20],inputs[argc];float phi_1,phi_2,phi_3;int i,j,k;int split=0;

void runge4(double x, double y[], double step);

if (argc<=3){cout << "\n\n\nYou need to include command line arguments!!\n" << endl;cout << "\nExample: ./a.out <phi_1> <phi_2> <phi_3> <friction> <1=split>!!\n\n\n" << endl;return 0;

}else{

sscanf(argv[1], "%f", &phi_1);sscanf(argv[2], "%f", &phi_2);sscanf(argv[3], "%f", &phi_3);sscanf(argv[4], "%f", &u);if (argc==6){sscanf(argv[5], "%d", &split);}

}

y[0]= phi_1;y[1]= 0.0;y[2]= phi_2;y[3]= 0.0;y[4]= phi_3;y[5]= 0.0;

for (j=1; j*DELTA_T <= T_MAX ;++j) {t=j*DELTA_T;runge4(t, y, DELTA_T);if (split == 1){

printf("%lf\t", j*DELTA_T);printf("%lf\t", y[0]+1);printf("%lf\t", y[2]);printf("%lf\t", y[4]-1);

printf("\n");}else{printf("%lf\t", j*DELTA_T);printf("%lf\t", y[0]);

printf("%lf\t", y[2]);printf("%lf\t", y[4]);

printf("\n");}

}

return 0;}//end of main

double f(double x, double y[], int i) {

double S1 = (D+L*(sin(y[2])-sin(y[0])));

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double S2 = (D+L*(sin(y[4])-sin(y[2])));

double f1=((K*S1*sin(y[0])) - (m*g*cos(y[0])))*u;double f2=((K*(S2-S1)*sin(y[2])) - (m*g)*cos(y[2]))*u;double f3=((-K*S2*sin(y[4])) - (m*g)*cos(y[4]))*u;

if (i==0) return(y[1]);

if (i==1) return(((K*S1*cos(y[0])) +f1+ (m*g*sin(y[0])))/(m*r));

if (i==2) return(y[3]);

if (i==3) return(((K*(S2-S1)*cos(y[2])) +f2+ (m*g)*sin(y[2])/(m*r)));

if (i==4) return(y[5]);

if (i==5) return(((-K*S2*cos(y[4])) +f3+ (m*g)*sin(y[4]))/(m*r));

}//end of f

void runge4(double x, double y[], double step) {double h=step/2.0;double t1[N], t2[N], t3[N];double k1[N], k2[N], k3[N],k4[N];int i;

for (i=0;i<N;i++) t1[i]=y[i]+0.5*(k1[i]=step*f(x, y, i));for (i=0;i<N;i++) t2[i]=y[i]+0.5*(k2[i]=step*f(x+h, t1, i));for (i=0;i<N;i++) t3[i]=y[i]+(k3[i]=step*f(x+h, t2, i));for (i=0;i<N;i++) k4[i]=step*f(x+step, t3, i);

for (i=0;i<N;i++) y[i]+=(k1[i]+2*k2[i]+2*k3[i]+k4[i])/6.0;return;

}//end of runge4

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8 References Engineering Mechanics Statics 2nd edition, Pytel & Kiusalaas, 1999.

Engineering Mechanics Dynamics 2nd edition, Pytel & Kiusalaas, 1999.

Physics for scientists and engineers 4th edition, Paul Tipler, 1999.

Honeywell Plastics: Nylon 6 Resins, http://www.honeywell-plastics.com/ed/solutions/7 16.html

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