PHY4222/Spring 08: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #11: Solutions
due by 12:50 p.m. Wed 04/24
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without such a
check, no more than 75% of the credit will be given even for an otherwise correct solution.
1. Show that the horizontal displacement of a simple pendulum of length ` satisfies the following non-linear equation
g
g
ẍ = − x + 3 x3 ,
`
2`
where the terms of order x3 have been kept. [33 points]
Solution:
The origin is at the pivot. Equations of motion:
mẍ = −Tx
mÿ = −Ty + mg,
where Tx and Ty are the projections of the tension. Balance of forces along the string
T = mg cos θ,
where θ is the angle the string makes with the vertical.
Tx = T sin θ = mg cos θ sin θ = mg sin θ
p
1 − sin2 θ
From geometry, sin θ = x/`, where ` is the length of the string. Now,
r
x 2
g
1 x 2
g
g
g
≈− x 1−
ẍ = − x 1 −
= − x + 3 x3 .
`
`
`
2 `
`
2`
2. A bob of mass m is attached by a weightless string of length b to the end of a spring (spring constant k). The
spring slides along a frictionless rail, as shown in the Figure. Assuming small oscillations, find the frequencies
of the normal modes of this system. [33 points]
Solution
Cartesian coordinates of the bob
x = a(t) + b sin θ
y = b cos θ
a(t) is the elongation of the spring. Kinetic energy
T =
1
1 2
mȧ2 + m bθ̇ + mȧbθ̇ cos θ.
2
2
In the harmonic approximation,
1
mȧ2 +
2
1
= mȧ2 +
2
T =
1 2
m bθ̇ + mȧbθ̇
2
1
m (ṙ)2 + mȧṙ,
2
2
FIG. 1: Problem 2
where r = bθ. Potential energy (dropping the constant in the harmonic approximation
U = −mgb cos θ +
θ2
a2
k 2
a ≈ −mgb + mgb + k
2
2
2
g
a2
= m r2 + k .
b
2
Matrix of masses
T̂ =
m m
m m
Û =
m gb
.
Matrix of spring constants
k
.
Characteristic equation
Det ω 2 T̂ − Û = 0 →
mω 2
mω 2 − m gb
= 0→
Det
mω 2
mω 2 − k
g
mω 2 − m
mω 2 − k = m2 ω 4
b
Because the ω 4 cancels out, there is only one root for ω 2 :
ω2 =
3. T&M 12.18. [34 points]
(g/b) (k/m)
.
g/b + k/m
12-18.
x
M
θ
b
m
y1
x1
x1 = x + b sin θ ; x& 1 = x& + bθ& cos θ
y1 = b − b cos θ ;
y& 1 = bθ& sin θ
Thus
T=
=
1
1
Mx& 2 + m x& 12 + y& 12
2
2
(
)
1
1
Mx& 2 + m x& 2 + b 2 θ& 2 + 2b x&θ& cos θ
2
2
(
)
U = mgy1 = mgb (1 − cos θ )
For small θ, cos θ
1−
θ2
2
. Substituting and neglecting the term of order θ 2θ& gives
T=
1
1
( M + m) x& 2 + m b 2θ& 2 + 2b x&θ&
2
2
U=
mgb 2
θ
2
(
)
Thus
⎡ M + m mb ⎤
m=⎢
mb 2 ⎥⎦
⎣ mb
0 ⎤
⎡0
A=⎢
⎥
⎣0 mgb ⎦
We must solve
−ω 2 ( M + m)
−ω 2 mb
=0
mgb − ω 2 mb 2
−ω 2 mb
which gives
ω 2 ( M + m) (ω 2 mb 2 − mgb ) − ω 4 m2 b 2 = 0
ω 2 ⎡⎣ω 2 Mb 2 − mgb ( m + M ) ⎤⎦ = 0
Thus
ω1 = 0
ω2 =
∑(A
j
jk
g
( M + m)
mb
)
− ω r2 m jk a jr = 0
Substituting into this equation gives
( k = 2, r = 1)
a21 = 0
a12 = −
bm
a
( m + M ) 22
( k = 2, r = 2)
Thus the equations
x = a11 η1 + a12 η2
θ = a21 η1 + a22 η2
become
x = a11 η1 −
mb
a η
( m + M ) 22 2
θ = a22 η2
Solving for η1 , η2 :
η2 =
θ
a22
x+
n1 =
bm
θ
(m + M)
a11
n1 occurs when n2 = 0; or θ = 0
n2 occurs when n1 = 0; or x = −
bm
θ
(m + M)