MIT 8 03 Fall 2005 Analysis of the Driven Triple Pendulum We wish to study the behavior of three pendulums each of mass m and length L in the con guration shown below under the in uence of a variable displacement 0 cos t at the top end We rst nd the normal mode frequencies by considering the case in which the system is un driven i e t 0 We then proceed to nd the amplitudes of oscillations of the pendulums as a function of the driving frequency We begin by setting a convenient coordinate system and labeling the relevant variables 1 L x1 L 2 x2 3 L x3 The gure below shows the forces acting on the individual pendula T1 T2 1 0 mg 11 00 2 T3 1 0 T3 mg mg For small angles T1 3mg T2 2mg and T3 mg If all three angles are zero and the system is at rest then this follows immediately The acceleration in the y direction is negligible for small angles Note that sin 1 x1 L sin 2 x2 x1 L and sin 3 x3 x2 L Then the equations of motion for the pendulums for small oscillations are mx 1 mx 2 mx 3 T2 sin 2 T1 sin 1 T3 sin 3 T2 sin 2 T3 sin 3 2mg x2 x1 L 3mg x1 L mg x3 x2 L 2mg x2 x1 L mg x3 x2 L where we have made the small angle approximations sin and cos 1 Rewriting the equations of motion in terms of 0 g L and rearranging terms x 2 x 1 02 5x1 2x2 2 0 2x1 3x2 x3 x 3 02 x3 x2 3 02 0 0 We use the normal mode anzatz xi Ci cos t where i 1 2 3 Note that x i 2 xi Then the set of coupled di erential equations xi unknown becomes a system of linear algebraic equations Ci unknown C1 2 5C1 02 2 02 C2 C2 2 2C1 02 3 02 C2 02 C3 C3 2 02 C2 02 C3 1 3 02 0 0 0 We can write these equations in the compact form Ax b where A is the matrix of coe cients x is the vector of amplitudes unknowns and b is the source vector In matrix form this means 2 02 0 3 02 0 C1 5 02 2 2 02 3 02 2 02 C2 0 2 2 2 0 0 0 C3 0 We now wish to nd the normal mode frequencies in the source free case i e Ax 0 Thus det A 6 06 18 04 2 9 02 4 6 0 I Igor Sylvester solved this equation using MAPLE The exact solutions are too complicated to present here Instead I give the following approximations for the normal mode frequencies 1 2 3 0 6448 0 1 5147 0 2 508 0 You may ask what happened to the other three roots Since the determinant is a polynomial of degree six it should have six roots In fact the complete solution to det A 0 is given by 2 12 22 32 Since negative frequencies do not add new physics we use only the positive values of Since we know the roots of det A we can write it in the convenient form det A 12 2 22 2 32 2 The relative amplitudes of oscillation at the normal modes are 2 02 02 2 C3 2 04 C2 C1 2 04 4 02 2 4 C1 2 04 4 02 2 4 triple tar triple tar The approximate values are C2 C2 2 29 C1 C1 1 2 C3 C3 3 92 C1 C1 1 2 1 35 1 05 C2 C1 3 C3 C1 3 0 65 0 12 We now re consider the forced case i e Ax b and solve for the amplitudes Ci using Cramer s rule 3 02 0 2 02 0 02 0 3 02 2 2 2 0 0 0 2 3 02 2 04 4 02 2 4 C1 0 2 det A 1 2 22 2 32 2 5 02 2 3 02 0 0 2 2 02 0 0 2 2 0 0 0 6 04 0 0 0 2 C2 det A 1 2 22 2 32 2 5 02 2 2 02 3 02 0 2 2 2 02 3 0 0 2 0 0 0 6 06 0 2 C3 det A 1 2 22 2 32 2 A graph of the amplitudes is shown on the next page 2 Triple Pendulum 6 4 C 0 2 0 2 4 6 0 0 5 1 1 5 0 2 2 5 3 3 5 C1 0 top mass closest to driver C middle mass 2 0 C bottom mass 3 0 It is interesting to note that the rst mass is stationary i e C1 0 if 2 2 2 2 0 only real solutions 1 849 0 764 0 Even more interesting is that the second mass is stationary if 0 If you take the triple pendulum and drive it at a frequency equal to that of a simple pendulum of mass m and length L then the second mass will not move You SHOULD ask yourself the question How on Earth can the two masses 2 and 3 move if the upper mass 1 does not move at all The answer is simple it is not possible It is only possible in our dream world of zero damping In the presence of damping no matter how little the peculiar state is unstable You will be able to go through that special state by varying omega but you cannot stop there I strongly recommend that you read Professor Lewin s own words at the end of the solutions to Problem 3 5 3