PHY6426/Fall 2007 CLASSICAL MECHANICSHOMEWORK ASSIGNMENT #8: solutions

due by 9:35 a.m. Mon 10/22Instructor: D. L. Maslov

maslov@phys.ufl.edu 392-0513 Rm. 2114

Please help your instructor by doing your work neatly.

1. Goldstein, Problem 6.12

Solution: η1.2 are displacement of the masses from their equilibrium positions. Lagrangian

L =1

2mη2

1+

1

2mη2

2−

1

2

(

kη2

1+ kη2

2+ 3k (η1 − η2)

2)

.

Equations of motion

d

dt

∂L

∂η1,2

=∂L

∂η1,2

or

mη1 = −kη1 − 3k (η1 − η2)

mη2 = −kη2 + 3k (η1 − η2)

Looking for solutions in a form of η1,2 = A1.2eiωt gives

−ω2A1 = −k

mA1 − 3

k

m(A1 − A2)

−ω2A2 = −k

mA2 + 3

k

m(A1 − A2)

which is equivalent to finding the roots of the equation∣

∣

∣

∣

∣

∣

∣

∣

ω2 − 4k/m 3k/m3k/m ω2 − 4k/m

∣

∣

∣

∣

∣

∣

∣

∣

= 0.

(

ω2 − 4k/m)2 − (3k/m)2 = 0

ω2

1= 7k/m; ω2 =

√

k/m.

2. Find the period of oscillations for a quartic one-dimensional oscillator

U (x) =1

4ax4.

Solution: energy conservation

1

2mx2 + U (x) = E.

Turning points, where x = 0, are at x = ±A. At these points, U(x) = E, so that E = 1

4aA4. Period

T = 4

∫ A

0

dx√

(2/m) (E − U (x))

= 2√

2m

∫ A

0

dx√

1

4aA4 − 1

4ax4

= 4

√

2m

a

1

A2

∫ A

0

dx1

√

1 − (x/A)4

= 4C

√

2m

a

1

A,

2

where

C =

∫ 1

0

dy1

√

1 − y4=

1

4B

(

1

4,1

2

)

≈ 1.31 . . .

with B (x, y) = Γ (x) Γ (y) /Γ (x + y) being a beta-function. Finally,

T = 7.42

√

m

a

1

A.

Note that the period decreases with increasing amplitude.

3. Using the perturbation theory, find the eigenfrequencies and eigenmodes of a one-dimensional oscillator withpotential energy

U (x) =1

2kx2 +

1

4ax4

in the limit a → 0. Which dimensionless parameter controls the convergence of the perturbation theory?

Solution: Equation of motion

mx = −kx − ax3

or

x + ω2

0x = −αx3 →ω2

0

ω2x + ω2

0x = −αx3 −

(

1 −ω2

0

ω2

)

x

where ω2

0= k/m and α = a/m. Seek for a solution in a form

x = A cosωt + x1 . . .

ω = ω0 + ω1 + . . .

To leading order in α, this gives

x1 + ω2

0x1 = −αA3 cos3 ωt + 2ω0ω1A cosωt.

Transform cos3 ωt as

cos3 ωt = cosωt cos2 ωt = cosωt1

2(1 + cos 2ωt)

=1

2cosωt +

1

2cosωt cos 2ωt =

1

2cosωt +

1

4(cos ωt + cos 3ωt)

=3

4cosωt +

1

4cos 3ωt.

The eq-n of motion now take the form

x1 + ω2

0x1 = −

3

4αA3 cosωt −

1

4αA3 cosωt + 2ω0ω1A cos ωt.

The term oscillating with frequency ω in the RHS looks like a force almost at resonance with the originalfrequency. This resonance is unphysical as our anharmonic oscillator is an isolated system. To avoid theresonance, we require that the coefficient of the cosωt term vanishes. This gives

ω1 =3

8α

A2

ω0

.

Finally, the equation for x1 reduces to

x1 + ω2

0x1 = −

αA3

4cos 3ωt.

3

The solution is

x1 = −αA3

32ω2

0

cos 3ωt.

Collecting everything together, the lowest order solution is

x = A cosωt +αA3

32ω20

cos 3ωt

ω = ω0 +3

8α

A2

ω0

= ω0

(

1 +3

8α

A2

ω20

)

.

The dimensionless parameter controlling the convergence of the perturbative expansion is 3

8αA2

ω2

0

.