Normal Modes
Problem 11.26 (a) Normal modes
Consider the system in …gure in which a bead of mass m is threaded on a
frictionless wire hoop of radius R and mass m. The hoop is suspended at point
A and is free to swing in its own vertical plane.
A frictionless hoop of mass m and radius R supports a bead of mass m which
is free to travel around the hoop. The hoop is suspended from point A and is
free to swing in its own vertical plane.
Using the suggested generalized coordinates in the …gure, the Cartesian coordinates (y being measured downward) of the bead are
x = R sin
1
+ R sin
2
and y = R cos
1
+ R cos
2:
(1)
The corresponding velocities are
x = R cos
1 1
+ R cos
2 2
and y =
R sin
R sin
1 1
2 2:
(2)
Hence the kinetic energy of the bead is
Tb =
2
2
1
mR2
2
1
+
2
+ 2 cos (
1
2)
(3)
1 2
From the parallel axis theorem the momentum of inertia of the hoop about A
is I = 2mR2 , hence the kinetic energy of the hoop is
2
Th = mR2
1:
(4)
The gravitational potential energy of the system is
U=
mgR (2 cos
1
+ cos
2) :
(5)
In the limit of small oscillations the Lagrangian (to within some meaningless
constants) is
L=
1
mR2 3
2
2
1
2
+
2
+2
1 2
1
1
mgR 2
2
2
1
+
2
2
:
(6)
The equations of motion are
1
:
2
:
@L
=
@ 1
@L
=
@ 2
2mgR
mgR
1
2
= mR2 3
= mR2
1
1
+
+
(7a)
2
:
2
(7b)
We can now write the the mass and spring matrix as
M=
3
1
1
1
and
K=
g
R
2
0
0
1
2
0
= ! 2o
0
1
;
(8)
where ! 2o = g=R: With these matrices the eigenvalue equation is
K ! 2 M a = 0;
(9)
which has a solution if and only if
det K ! 2 M = det
3! 2
2! 2o
!2
!2
= 2! 4
5! 2 ! 2o + 2! 4o = 0; (10)
! 21 = ! 2o =2 and ! 22 = 2! 2o :
(11)
!
2
! 2o
or
The eigenvalue equation for the …rst eigenvector is
2
3=2
1=2
1=2 1 1=2
a1
a2
a1
1=2
1=2
=
a2
=
1=2
1=2
a1
a2
=0
0 ! a1 = a2 :
(12)
The eigenvalue equation for the …rst eigenvector is
2
6
2
2 1 2
a1
a2
2a1 + a2
=
4
2
2
1
=
0 ! a1 =
a1
a2
=0
a2 =2:
(13)
The normal modes are
= A1
1
1
cos
g
p t
2R
1
and
= A2
1
2
cos
p
2g
t
R
2
!
:
(14)
In the …rst mode the hoop and the bead are oscillating in phase with equal
amplitudes, the bead and hoop oscillate together. In the second mode the hoop
and bead oscillate exactly out of phase with the amplitude of 2 being twice of
1:
(b) Normal coordinates
For this case expanding in terms of its eigenvectors yields
= p1
2
1
1
+ p2
5
2
1
2
:
The equations of motion for
were
3
1
+
2
=
2! 2o
1
+
2
=
! 2o
where ! 2o = g=R: Substituting for
2;
in terms of the normal coordinates yields
4 p1 + p2
2
5
=
2! 2o
2 p1
2
=
! 2o
p2
5
1;
p1 + p2
2
5
p1
2
2 p2
5
;
:
Simply adding these two equations yields
6 p1 =
2
3! 2o p1 !
2
1
=
! 2o
2
1:
Multiplying the second EOM by 2 and subtracting them yields
3 p2 =
5
6! 2o p2 !
5
2
=
2! 2o 2 :
Again we see that the normal coordinates are decoupled and oscillate at their
normal mode frequencies.
To solve for the normal coordinates we simply invert the expansion
1
=
p1 + p2 ;
2
5
2
=
p1
2
2 p2 ;
5
which leads to
p
1= p5
2= 5
2
p
p
=
1=p2 1= p5
det
1= 2
2= 5
p
1=p2
1
det
1= 2
2
p
p
=
1=p2 1= p5
det
1= 2
2= 5
det
1
1
=
=
1
2
p
1=
p
p
5
2
2= 5
p
=
(2
3
3= 10
p
5
(
3
3
1
2) :
1
+
2) ;