Exercise 4: Coupled Oscillations
Due at 5pm on Tuesday, September 28, in Matt’s mailbox. (No separate problem set this week.)
Review of lecture material, Monday/Wednesday Sept 20/22
> restart;
For two masses 1 and 2, connected by three springs a, b, and c, the differential equations of
motion for the three masses are
> eq1:=m1*diff(x1(t),t,t)=-ka*x1(t)-kb*(x1(t)-x2(t));
 ∂2

eq1 := m1  2 x1( t )  = −ka x1( t ) − kb ( x1( t ) − x2( t ) )
 ∂t

> eq2:=m2*diff(x2(t),t,t)=-kc*x2(t)+kb*(x1(t)-x2(t));
 ∂2

eq2 := m2  2 x2( t )  = −kc x2( t ) + kb ( x1( t ) − x2( t ) )
 ∂t

We simplifed this by making both masses the same value as well as the three spring constants, and
k
2
also by defining ω0 =
m
> m1:=m:
> m2:=m:
> ka:=m*omega0^2:
> kb:=m*omega0^2:
> kc:=m*omega0^2:
> eq1:=expand(eq1/m);
eq1 :=
∂2
∂t
2
x1( t ) = −2 ω0 x1( t ) + ω0 x2( t )
2
2
> eq2:=expand(eq2/m);
eq2 :=
∂2
∂t
2
x2( t ) = −2 ω0 x2( t ) + ω0 x1( t )
2
2
Into these, we substituted x1 = A1 e( i ω t ) and x2 = A2 e( i ω t ). We then have the two algebraic
equations for A1 and A2:
> eq1a:=(omega^2-2*omega0^2)*A1+omega0^2*A2=0;
eq1a := ( ω − 2 ω0 ) A1 + ω0 A2 = 0
> eq2a:=omega0^2*A1+(omega^2-2*omega0^2)*A2=0;
2
2
2
eq2a := ω0 A1 + ( ω − 2 ω0 ) A2 = 0
These are two equations for A1 and A2 which have the only solution A1=A2=0, unless they are
actually the same equation. So, to avoid the trivial solution, we make them the same equation by
2
2
2
multiplying the first by ω0 and the second by ω − 2 ω0 and then forcing the coefficients of A2
2
2
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2
to be the same. This gives us an equation to solve for ω . The result is the frequencies of the two
"normal modes" of this coupled oscillator.
> sol:=solve((omega^2-2*omega0^2)^2=omega0^4,omega^2);
2
sol := ω0 , 3 ω0
Either of these two solutions gives us the same equation from the two we wrote down for A1 and
A2. The first solution implies that
> simplify(subs(omega^2=sol[1],eq1a));
2
2
−ω0 A1 + ω0 A2 = 0
This means that A1=A2. The two masses move together with the same amplitude. The spring in
between them never stretches. The second solution implies that
> simplify(subs(omega^2=sol[2],eq1a));
2
2
ω0 A1 + ω0 A2 = 0
This means that A1=-A2. The two masses move against each other, but again with the same
amplitude. We can see what these normal modes actually mean in terms of the equations of
motion by defining new quantities x(t) and y(t) and substituting them into the differential
equations for x1(t) and x2(t).
> xyx1x2:={x(t)=(x1(t)+x2(t))/2,y(t)=x1(t)-x2(t)}:
> x1x2xy:=solve(xyx1x2,{x1(t),x2(t)}):
> neweqs:=subs(x1x2xy,{eq1,eq2}):
> solve(neweqs,{diff(x(t),t,t),diff(y(t),t,t)});
2
{
∂2
2
y( t ) = −3 ω0 y( t ),
2
∂2
x( t ) = −ω0 x( t ) }
2
∂t
∂t
These differential equations are much simpler, and show that the coordinate y(t), which represents
the distance between the two masses, oscillates with a frequency that is 3 times faster than the
coordinate x(t), which represents the center of mass of the two masses.
The actual motion of the two masses can be solved directly in MAPLE. Take the case where mass
1 starts from rest at x1=A, while mass two starts from rest at equilibrium. Then we get
> paths:=dsolve({eq1,eq2,x1(0)=A,x2(0)=0,D(x1)(0)=0,D(x2)(0)=0},{
x1(t),x2(t)});
2
2
paths :=
1
1
1
1
{ x2( t ) = cos( ω0 t ) A − cos( 3 ω0 t ) A, x1( t ) = cos( ω0 t ) A + cos( 3 ω0 t ) A }
2
2
2
2
> A:=5: omega0:=1:
> plot([rhs(paths[1]),rhs(paths[2])],t=0..4*2*Pi/omega0,color=[gr
een,magenta]);
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4
2
0
2
4
6
8
10
12
t
14
16
18
20
22
24
–2
–4
The motions are clearly complicated. However, if we plot instead x(t) and y(t), things look a lot
simpler.
> xypaths:=solve(subs(paths,x1x2xy),{x(t),y(t)}):
> plot([rhs(xypaths[1]),rhs(xypaths[2])],t=0..4*2*Pi/omega0,color
=[blue,red]);
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4
2
0
2
4
6
8
10
12
t
14
16
18
20
22
24
–2
–4
That is, the center of mass (blue) and the difference between the positions (red) oscillate as sine
waves all by themselves. It is also possible to calculate the kinetic energy of the two masses
individually from the solution, and also the potential energy in the three springs.
> assign(paths);
> K1:=m*diff(x1(t),t)^2/2:
> K2:=m*diff(x2(t),t)^2/2:
> U:=k*x1(t)^2/2+k*x2(t)^2/2+k*(x1(t)-x2(t))^2/2:
> U:=subs(k=m*omega0^2,U):
> m:=1:
> plot([K1,K2,U,K1+K2+U],t=0..4*2*Pi/omega0,color=[green,magenta,
yellow,black]);
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24
22
20
18
16
14
12
10
8
6
4
2
0
2
4
6
8
10
12
t
14
16
18
20
22
24
The sum of the three energies is indeed a constant.
Exercise: Initial conditions corresponding to normal modes
Use MAPLE to solve the coupled differential equations for x1( t ) and x2( t ) for
two new sets of initial conditions. For one of these, start the oscillator off so that
it excites only one of the normal modes, and the other using the other set of
initial conditions. In each case, show that the behavior of x( t ) and y( t ) is what
you expect.
Finally, we pointed out that there are standard techniques for finding the frequencies and "normal
coordinates" (which we called x(t) and y(t)) using matrix techniques. The procedure is called the
"eigenvalue" method, and is done withing MAPLE using the "linalg" package. For this case,
> restart;
> Omega:=matrix([[-2*omega0^2,omega0^2],[omega0^2,-2*omega0^2]]);
−2 ω02
Ω := 
2
 ω0
2
ω0 
2
−2 ω0 
> with(linalg):
Warning, new definition for norm
Warning, new definition for trace
> eigenvalues(Omega);
−3 ω0 , −ω0
2
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2
> eigenvectors(Omega);
[ −ω0 , 1, { [ 1, 1 ] } ], [ −3 ω0 , 1, { [ 1, -1 ] } ]
2
2
The eigenvalues are the negatives of the ω values for the normal modes, and the eigenvectors
corresponding to each of these two eigenvalues (the two component vector in the last element of
each of the two lists returned from "eigenvectors") are the relative values of the "A" coefficients,
and tell us the way to constuct the normal coordinates. As shown here, the first eigenvalue has the
two masses oscillating with the same amplitude but opposite to each other, and the second has the
two oscillating at the same amplitude and in phase, just as we found with the detailed algebra
above.
2
Extension Exercises
1) Two masses are connected to each other by a spring with spring constant α,
and each is connected to a fixed wall on either side by spring with spring
constant k. (This is the same as the case worked in class, but the spring in the
middle no longer has the same spring constant as the the two on the outside.)
• Write the equations of motion and show that the two normal modes have frequency ω where
k
2α
2
2
2
2
2
2
2
.
ω = ω0 and ω = ω0 + β where ω0 = and β =
m
m
• Show that the equations decouple if we write x =
x1 + x2
2
and y = x1 − x2.
• Find the solutions x1 and x2 where mass 1 starts from rest at position x0 and mass 2 starts from
rest at equilibrium.
• Make a plot of the two solutions for t = 0 to
200 π
ω0
using
β
ω0
= .2.
• Plot the kinetic energies of the two masses for the same conditions. Briefly describe what is
going on here in words.
2) Now consider the case of two unequal masses m1 = m( 1 − µ ) and
m2 = m( 1 + µ ), where µ is postive but less than one, connected by three
identical springs with spring constant k to fixed walls on the end. Find the
k
2
frequencies of the normal modes, in terms of µ and ω0 = , and show that
m
they reduce to the proper values when µ = 0. Also, take the limit as µ
approaches one (from the "left" - see the "limit" help topic) and interpret the
resulting frequencies. (Hint: If two identical springs with spring constant k are
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connected in series, the effective spring constant of the combination is
k
2
.)
3) A mass m hangs vertically from a spring with spring constant k. A second,
identical mass hangs from the first mass, also by a spring with spring constant
k. Find the frequencies of the normal modes, the normal coordinates which
decouple the equations, and describe the motion of the normal modes.
4) Three identical masses m move horizontally on a frictionless track. They are
connected to each other by identical springs with spring constant k and the two
on the end are connected to fixed walls, again by identical springs with spring
constant k. Find the frequencies of the normal modes, and describe the motion
of the normal modes. To describe the motion of the normal modes, use dsolve
to solve the differential equations, using initial conditions with each of the three
masses at rest, and where they start at values dictated by the eigenvectors. Plot
the paths of the three masses for two periods of the eigenfrequency and
confirm that the motions are indeed sinusoidal with the correct period.
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