PHYS 611, Fall 2025

Theoretical Mechanics

Homework Set 1. - TA Solution

Thomas (Tom) C. Gorordo

1. Damped oscillator

F = -m ((ω0)^2)x[t] - m γ x'[t];
DHODE = m x''[t]== F

(a)

Xa[t_]:= Δ (E^(I ω t));
DHODE/.{x->Xa}//Simplify

eliminating the common factor as nonzero, find a quadratic condition that:

CONDa = (-I γ ω+(ω^2)-((ω0)^2))==0;

(b)

Solve[CONDa, ω]

from which we can observe that solutions will be pure imaginary if

CONDb = -(γ^2)+4 ((ω0)^2) <= 0;

(c)

Xc[t_]:= ap (E^(-kp t))+am (E^(-km t))

Since we’re solving a 2nd order homogeneous ODE, we know we can form a general solution as a superposition of the two solutions found in (a), so we can specifically take

kp = -I ω/. Solve[CONDa, ω][[1]]//Expand
km = -I ω /.Solve[CONDa, ω][[2]]//Expand

where the condition found in (b) implies we are in an overdamped (or critically damped) regime.
We can check this indeed yields a solution:

DHODE/.{x->Xc}//Simplify

noticing that the second term in these expressions is the hinted at frequency parameter

ωw == (1/2) I (Sqrt[-(γ^2)+4 ((ω0)^2)]);

it will simplify our further manipulations to write

kp = (γ/2)+ωw;
km = (γ/2)-ωw;

(d)

The Cauchy boundary conditions

CONDd = {x[0] == X0, x'[0] == V0};

imply coefficients

COEFFSd = Solve[CONDd/.{x->Xc},{ap, am}][[1]]//Expand

so the solution takes the form

Xd[t_]:= Xc[t]/.COEFFSd

Xd[t]

(e)

Xe[t_]:=Limit[Xd[t],ωw->0]
Xe[t]

2. Why the harmonic oscillator is special

U[x_]:=A ((Abs[x])^n)

(a)

The turning points occur when the whole energy is potential

Xturn = x[t]/.Solve[EE == U[x[t]], x[t]]

xturn = Xturn[[2]]

(b)

T = ∫dt = ∫((dt)/(dx))dx

where

DEFb = 
EE == (1/2)m (((x'[t]))^2)+Hold[U[x[t]]];

so dt/dx is

(1/(x'[t]))/.Solve[DEFb, x'[t]][[2]]

thus, we aim to integrate

√((m/(2 EE)))(∫_(Xturn[[1]])^(Xturn[[2]]))(1/(√(1-((U[x[t]])/EE))))dx

breaking down the integral further, notice that

((U[x[t]])/EE) == ((U[x[t]])/(U[xturn])) == ((Abs[((x[t])/xturn)])^n)

and the integral can be rewritten

√((m/(2EE)))(∫_0^xturn)(1/(√1-((Abs[x/xturn])^n)))dx

which is nondimensional under the substitution w = x / xturn

T= 2 Assuming[{n>0, n∈Integers}, xturn√((m/(2EE)))(∫_0^1)(1/(√(1-(u^n))))\[DifferentialD]u]

2 Assuming[{n>0, n∈Integers}, (∫_0^1)(1/(√(1-(u^n))))\[DifferentialD]u]

(c)

We didn’t need to evaluate the integral to do this next part, because it will vanish in a ratio:

(1/T)D[T,EE]//Simplify//Expand

3. Linear two-spring oscillator

(a)

SOLa = Solve[0 == ((m(x1[t] + x2[t]) + M X[t])/(2m + M)),X[t]]

(b)

EOM = m{{x1''[t]},{x2''[t]}}=={{-k, 0},{0, -k}}.{{x1[t]},{x2[t]}} + k{{X[t] - d},{X[t]+d}};

EOM/.SOLa//Expand

(c)

CONDc = EOM/.SOLa/.{x1''[t]->0, x2''[t]->0}//Simplify

Solve[CONDc, {x1[t], x2[t]}]

(d)

Xd1[t_]:=a1 (E^(I ω t))-d
Xd2[t_]:=a2 (E^(I ω t))+d

Setting a1 = a2 = a

EOM/.SOLa/.{x1->Xd1, x2->Xd2}/.{a1->a,a2->a}//Simplify

we can read off a sufficient condition for the frequency of a nontrivial solution

Solve[(k (2 m+M)-m M ωp2) ==0, ωp2]

and the a1 = -a2 = a ansatz proceeds similarly

EOM/.SOLa/.{x1->X1, x2->X2}/.{a1->a,a2->-a}//Simplify

Solve[ (k-m ωm2)==0,ωm2]

wp2 = wm2 (((2m + M))/M)

(e)

X1[t_]:= -d +am (E^(I ωm t))+Conjugate[am](E^(-I ωm t))+ap (E^(I ωp t))+Conjugate[ap] (E^(-I ωp t))
X2[t_]:=d - am (E^(I ωm t))-Conjugate[am](E^(-I ωm t))+ap (E^(I ωp t))+Conjugate[ap](E^(-I ωp t))

CONDe = {X1[0]==-d,X2[0]==d,X1'[0]==0,X2'[0]==v}//Simplify

X1[t]/.Solve[CONDe,{ap, am, Conjugate[ap], Conjugate[am]}]//ExpToTrig
X2[t]/.Solve[CONDe,{ap, am, Conjugate[ap], Conjugate[am]}]//ExpToTrig

(f)

SOLa/.{x1->X1,x2->X2}/.Solve[CONDe,{ap, am, Conjugate[ap], Conjugate[am]}]//ExpToTrig//Simplify

(g)

(X2[t]-X1[t])/.Solve[CONDe,{ap, am, Conjugate[ap], Conjugate[am]}]//ExpToTrig//Simplify

where this frequency is smaller (see (d)).

(h)

Plot[Evaluate[{
X1[t],
X2[t],
X[t]/.SOLa/.{x1->X1,x2->X2},
X2[t]-X1[t]}
/.Solve[CONDe,{ap, am, Conjugate[ap], Conjugate[am]}]
/.{ωp->ωm √((((2m + M))/M))}/.{d->1,ωm->2, M->20m, v->1}//Simplify],
{t,0, 40π}, ImageSize->Large]

(i)

X''[t]==D[X[t]/.SOLa,{t,2}]/.Solve[EOM,{x1''[t],x2''[t]}]/.Solve[X[t]==(X[t]/. SOLa),{x1[t]}]//Simplify

δX''[t]==D[x2[t]-x1[t],{t,2}]/.Solve[EOM,{x1''[t],x2''[t]}]/.{x2[t]->δX[t]+x1[t]}//Simplify

whose frequencies are again seen to come from our physically intuited ansatz in (d).