PHY6426/Fall 07: CLASSICAL MECHANICS
FINAL EXAM: SOLUTIONS
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Total points: 120
Problem 1 A pendulum consists of a uniform rigid rod of length L, mass M, and of a snail of mass M/3 which can crawl
along the rod (Fig.1–turn the page over). The rod is pivoted at one end and swings in a vertical plane. Initially,
the snail is at the pivot-end of the rod and then it crawls slowly with constant speed v along the rod towards
the bottom end of the rod. Assume that the snail can be treated as a point mass.
a) Construct the Lagrangian for the rod-snail system and derive the equations of motion. [ 20 points]
When the snail has crawled distance `, the moment of inertia is
I=
1
1
M L2 + M `2 .
3
3
Lagrangian
L=
I θ̇2
M v2
1
1
+
+ M gL cos θ + M g` cos θ.
2
6
2
3
E.o.m.
∂L
d ∂L
=
dt ∂ θ̇
∂θ
1
1
d I θ̇ = − M gL sin θ − M g` sin θ
dt
2 3
dI
`
L
.
I θ̈ + θ̇ = −M g
+
dt
2
3
dI
2
= M ``˙
dt
3
θ̈ + 2
3
``˙θ̇
2L + `
=
−g
sin
θ
.
L2 + ` 2
L2 + ` 2
b) Find the frequency of small oscillations of the pendulum when the bug has crawled a distance ` along the
rod. Assume that the snail crawls so slowly that the change in ` over the period of oscillations can be
neglected (that is, discard the `˙ term in the equation and treat ` as a constant). [20 points]
Neglecting the term containing `˙ and expanding sin θ ≈ θ, we obtain
θ̈ = −gθ
3
2L + `
L2 + ` 2
and
ω=
s
g
3
2L + `
L2 + ` 2
Problem 2 Consider a damped linear oscillator described by the following equation of motion
q̈ + 2γ q̇ + ω 2 q = 0.
(1)
2
a) Assuming the Lagrangian of this system can be written as
1 2 1
2 2
mq̇ − mω q ,
L = f (t)
2
2
(2)
find such a function f (t) that the Euler equation reproduces the equation of motion (1). To define f (t)
uniquely, assume that f (0) = 1. [10 points]
d ∂L
∂L
=
dt ∂ q̇
∂q
d
(f (t) mq̇) = −f (t) mω 2 q
dt
f˙mq̇ + f mq̈ = − − f (t) mω 2 q
f˙
q̈ + q̇ = −ω 2 q.
f
The e.o.m. is reproduced, if
f˙
= 2γ.
f
Solving this equation with the initial condition f (0) = 1, we obtain
f (t) = exp (2γt) .
b) Construct the Hamiltonian H (q, p) using the Lagrangian from Eq. (2) with f (t) found in part a). [15
points]
∂L
= f (t) mq̇
∂ q̇
p
.
q̇ =
mf (t)
p =
The Hamiltonian
p2
1
+ f (t) mω 2 q 2
2mf (t) 2
2
1
p
+ e2γt mω 2 q 2 .
= e−2γt
2m
2
H = q̇p − L =
c) For the generating function
F2 (q, P, t) = exp (γt) qP,
find the transformed Hamiltonian K (Q, P, t) and show that K is conserved. [15 points]
∂F2
= eγt P
∂q
∂F2
Q =
= eγt q
∂P
q = e−γt Q
p =
K=H+
∂F2
= H + γqP eγt .
∂t
H is new variables
2
H = e−2γt
2
P2
1
1
(eγt P )
+ e2γt mω 2 e−γt Q =
+ mω 2 Q2
2m
2
2m 2
3
P2
1
+ mω 2 Q2 + γ e−γt Q P eγt
2m 2
P2
1
=
+ mω 2 Q2 + γQP.
2m 2
K =
K does not depend on time and is, therefore, conserved.
Problem 3 A non-linear oscillator has a potential given by
V (x) =
1 2 λ
kx − mx3 ,
2
3
where λ is a small parameter. Find the solution of the equation of motion to first order in λ, assuming that
x (0) = 0 and ẋ (0) = v0 . [40 points] Hint: apply the initial conditions at each step of the perturbation theory.
E.o.m.
ẍ + ω02 x = λx2 ,
where ω02 = k/m. Perturbation theory
x = x0 + x1 ,
where x0 = A cos ωt+B sin ωt. To zeroth order in λ, substitute x = x0 into the e.o.m. and neglect the non-linear
term:
−Aω 2 cos ωt + Aω02 cos ωt − Bω 2 sin ωt + Bω02 sin ωt = 0 → ω = ω0 .
Taking into account the initial conditions,
x0 =
v0
sin ω0 t.
ω0
To first order in λ, keep only the λx20 term in the RHS of the e.o.m.
ẍ1 + ω02 x1 = λx20 = λ
v0
ω0
2
sin2 ω0 t =
1
λ
2
v0
ω0
2
1
− λ
2
v0
ω0
2
cos 2ω0 t.
A general solution of the inhomogeneous linear differential equation is a sum of a fundamental solution of the
homogeneous equation and a particular solution of the inhomogeneous equation: x 1 = xf + xp . A fundamental
solution
xf = D cos ω0 t + E sin ω0 t.
We can look for a particular solution of the following form
xp = F + G cos 2ω0 t.
Substuting back into the equation, we find
2
v0
ω0
2
1
v0
−4Gω02 cos 2ω0 t + Gω02 cos 2ω0 t = − λ
cos 2ω0 t.
2
ω0
2
1
v0
−3Gω02 cos 2ω0 t = − λ
cos 2ω0 t.
2
ω0
2
v0
λ
G =
2
6ω0 ω0
F =
λ
2ω02
4
snail
l
L
FIG. 1:
2
2
v0
λ
v0
+
cos 2ω0 t
ω0
6ω02 ω0
2
2
v0
λ
v0
λ
+
=0
x1 (0) = 0 → D +
2ω02 ω0
6ω02 ω0
2
2λ v0
D = − 2
3ω0 ω0
x1 = D cos ω0 t + E sin ω0 t +
λ
2ω02
The zeroth order solution already satisfies the initial condition ẋ0 (0) = v0 . Therefore, the first-order correction
must satisfy ẋ1 (0) = 0, which gives E = 0. Finally,
2λ
x1 (t) = − 2
3ω0
v0
ω0
2
λ
cos ω0 t +
2ω02
v0
ω0
2
λ
+
6ω02
v0
ω0
2
cos 2ω0 t