PHYS 611, Fall 2025

Theoretical Mechanics

Homework Set 2. - TA Solution

Thomas (Tom) C. Gorordo

1. Dynamical matrix for coupled pendulums

(a)

Δ = -{{-k1 - k2, k2},{k2, -k1-k2}};
Δ//MatrixForm

u[t_] := {u1[t],u2[t]};

EOM = m  u''[t] == -Δ.u[t] //Simplify

(b)

{ω2s, vecs} = Eigensystem[(Δ/m)]

ωs =Sqrt[ω2s]

ws = vecs / √Length[vecs]

(c)

Λ = DiagonalMatrix[ωs];
m (Λ^2)//MatrixForm

ϕ =Table[Subscript[a, i, j],{i,2},{j,2}];

condc1 = {
(ϕ.u''[t] == - ((Λ^2)).ϕ.u[t]/.Solve[EOM,u''[t]][[1]]//Simplify),
Transpose[ϕ].ϕ == ϕ.Transpose[ϕ]==IdentityMatrix[2]
}//Simplify;
MatrixForm/@ (ϕ/.Solve[condc1, Flatten[ϕ]])

So, there are two choices and an arbitrary sign for the transformation (noticing that these amount
to moving into sum/difference coordinates, these amount to choosing which order you want the difference to be and which direction is positive)

MatrixForm/@EigenvalueDecomposition[Δ] 
Normalize/@Transpose[EigenvalueDecomposition[Δ][[1]]]//Transpose//MatrixForm(* It might be nice to take this to be canoncical *)

condc2 = {
(ϕ.u''[t] == - ((Λ^2)).ϕ.u[t]/.Solve[EOM,u''[t]][[1]]//Simplify),
Transpose[ϕ].ϕ == ϕ.Transpose[ϕ]==IdentityMatrix[2],
ϕ.ϕ==IdentityMatrix[2] (* but let's further specify we want involutory (symmetric) *)
}//Simplify;

Φ = (ϕ/.Solve[condc2, Flatten[ϕ]][[2]](* and pick the mostly positive one *));
Φ//MatrixForm

Φ.Δ.Φ//Simplify

(d)

V = (1/2)u[t].(Δ.u[t])//Simplify

F = m u''[t]/.Solve[EOM,u''[t]][[1]]//Simplify

-Grad[V,{u1[t],u2[t]}]==F//Simplify

(e)

KE = (1/2)Transpose[u'[t]].u'[t]
KE == (1/2)Transpose[Φ.u'[t]].Φ.u'[t]//Simplify

so it will be enought just to find the potential in the normal coordinates

V/.Solve[{q1[t],q2[t]}==Φ.u[t],u[t]]//Simplify

so

E1 = (1/2)((q1'[t])^2)+(1/2)k1 ((q1[t])^2);
E2 = (1/2)((q2'[t])^2)+(1/2)(k1+2k2) ((q2[t])^2);

D[E1, t]//Simplify
D[E2, t]//Simplify

but the factors on the right are each zero by the EOM!

2. Synchronicity in the linear chain

3. Dynamical matrix for masses on a ring