PHY6426/Fall 2007: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #9
due by 9:35 a.m. Wed 10/31
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
1. A charged bob (mass m, charge q > 0) is suspended in a uniform and static electric field E, as shown in Fig. 1
(left). The ratio of the electric force to gravity is arbitrary.
a) Find the frequency of small oscillations about the equilibrium position.[13 points]
T =
1 2 2
m` φ̇
2
Electric field produces a force on the bob
FE = qE
Corresponding potential energy
UE = −
Z
dxFE = −qEx = −qE` sin φ
Gravitational potential energy
Ug = −mg` cos φ
Alltogether,
L = T − U E − Ug =
1 2 2
m` φ̇ + mg` cos φ + qE` sin φ
2
d ∂L
∂L
= m`2 φ̈ =
= −mg` sin φ + qE` cos φ
dt ∂ φ̇
∂φ
The eqilibrium position of the bob is found from
φ̈ = 0 → mg` sin φ0 = qE` cos φ0
tan φ0 = qE/mg
φ0 = arctan (qE/mg) .
Using (1), we can re-write the RHS of the Euler’s equation as
cos φ0
∂L
= −mg` sin φ + qE` cos φ = −qE`
sin φ + qE` cos φ
∂φ
sin φ0
qE`
cos φ0
= −qE`
sin φ − cos φ = −
[sin φ cos φ0 − cos φ sin φ0 ]
sin φ0
sin φ0
qE`
= −
sin (φ − φ0 )
sin φ0
Expanding the right-hand side of the equation of motion about in θ = φ − φ0 (|θ| 1),we obtain
m`2 θ̈ = −
qE`
θ.
sin φ0
(1)
2
Therefore,
ω02 =
qE
m` sin φ0
A little bit of trigonometry and algebra leads to
1 + tan2 φ0 = 1/ cos2 φ0
p
tan φ0
1 − cos2 φ0 = p
sin φ0 =
1 + tan2 φ0
q
g
2
ω02 =
1 + (qE/mg)
`
!
g 2 qE 2 1/4
.
+
ω0 =
`
m`
e) Suppose that the pendulum is also driven by a periodic force F = F0 cos ωt parallel to the electric field.
What are the conditions for a resonance, if the non-linearity is taken into account? [20 points]
Expanding ∂L/∂φ further, we obtain
∂L
qE`
qE`
θ3
=−
sin θ ≈ −
θ−
∂φ
sin φ0
sin φ0
6
The θ3 term generates a fractional resonance at ω = ω0 /3.
2. The support point of a simple pendulum (mass m, length `) oscillates horizontally according to x 0 (t) = a cos ωt.
a) Construct the Lagrangian (remember to discard the total time derivatives, if needed). [15 points]
x = a cos ωt + ` sin φ
y = ` cos φ
ẋ2 + ẏ 2 = a2 ω 2 sin2 ωt + `2 φ̇2 − 2`φ̇ cos φ sin ωt.
The first term is the total time derivative (t.d.t) and hence can be discarded. The third term can be re-written
using the identity
φ̇ cos φ sin ωt =
d
(sin φ cos ωt) + ω sin φ cos ωt.
dt
Discarding the t.t.d and subtracting the potential energy, we obtain the Lagrangian
m 2 2 1
` φ̇ + ma`ω 2 sin φ cos ωt + mg` cos φ.
2
2
p
b) Which values of ω, apart from ω = ω0 = g/`, lead to a resonance in this system? [18 points]
L=
The eq-n of motion is
1
m`2 φ̈ = − ma`ω 2 cos φ cos ωt − mg` sin φ
2
1. Assume that a and φ are small. Expand the RHS to order
2
1
m`2 φ̈ = − ma`ω 2 1 − φ2 /2 cos ωt − mg`φ.
2
3
To order a0 (no external force), we have a free motion with frequency ω0 :
φ0 = α cos ω0 t.
The the first-order correcton (order a1 ) satisfies
1
m`2 φ̈1 = − ma`ω 2 1 − φ20 /2 cos ωt − mg`φ1 .
2
The new ”force” in the RHS of this equation ∝ a φ20 cos ωt ∝ a cos2 ω0 t cos ωt ∝ a cos ωt + a cos (2ω0 − ω) t +
a cos (2ω0 + ω) t. All of them lead only to the resonance at ω = ω0 . To order a2 , one needs to retain the cross
term φ0 φ1 obtained by expanding the square of φ0 + φ1 . This terms oscillates at combination frequencies ω0 ± ω.
The parametric resonance occurs for ω = 2ω0 .
2. A positive charge q is fixed at the end of a vertical rod. A positively charged bead (charge Q, mass m) can slide
along the rod without friction (see Fig. 1, right). The acceleration of gravity is g. Show that a periodic force of
frequency ω, applied along the rod, leads to a resonance if ω = ω0 , 2ω0 , 3ω0 , where ω0 is the frequency of small
oscilations about the equlibrium position of the bead. [34 points] [Hint: to get the resonance at 2ω 0 , think of a
parametric resonance.]
The potential energy of the bead
U=
qQ
+ mgx
4πε0 x
Equation of motion
mẍ = F (x) = −
qQ
∂U
=
− mg
∂x
4πε0 x2
The equlibirium position of the bead
x0 =
qQ
4πε0 mg
1/2
3
Expanding the force to order (x − x0 ) , we obtain
mẍ = (x − x0 ) F 0 (x0 ) +
1
1
2
3
(x − x0 ) F 00 (x0 ) + (x − x0 ) F 000 (x0 ) + . . .
2
6
or
ÿ + ω02 y =
1
1 2 00
y F (x0 ) + y 3 F 000 (x0 )
2
6
where
ω0 =
r
−F 0 (x0 )
m
By the equilibrium condition F 0 (x0 ) < 0.
The y 2 term produces a ”force” y0 y1 ∝ cos ωt cos ω0 t which leads to a resonance at ω = 2ω0 .
4
g
E
Q
q
FIG. 1: