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PHY354H1
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Erich Poppitz
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Lecture

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Physics

PHY354H1

Erich Poppitz

Winter

Description

Lecture 15: Coupled Oscillators & Normal Modes
▯ relevant reading: Morin 4.5
▯ Reminder: Fall break next Monday November 12 & Tuesday November
13 so no tutorials next week. But tutorial questions still available. TA’s
will have o▯ce hours during their Wednesday tutorial time.
Coupled Systems
▯ So far we have considered systems having only 1 free oscillation (i.e.
there is only 1 object that can move independently). This means that
the system has only a single natural frequenc0 \! " associated with it.
▯ !0depended onpthe restoring force in the system. Fpr example, for
springs: 0 = k=m, for small-angle pendulums, 0 = g=l.
▯ Real physical systems are generally capable of oscillating in many dif-
ferent ways, they may have more than 1 restoring force acting, they
may be coupled to other objects and hence a▯ected by their motion,
and they may have more than 1 natural frequency.
▯ However, there are still certain circumstances where the components of
the system oscillate at a single \natural frequency" and we would like to
▯gure out what this is. These frequencies are known as \characteristic
frequencies" or \normal modes" of the system. Once you know the
normal modes, it turns out you can write the general solution for the
system in terms of the normal modes.
Example: Two Coupled Pendulums
▯ Show demonstration of coupled pendulums.
1 ▯ Set-up and approximations:
{ Take 2 pendulums and a spring.
{ Make the width between the pendulums the exact width of the
unstretched spring (which we will call L), so when the pendulums
are vertical, there is no force due to the spring.
{ Assume the displacements of the pendulum bobs are small and
hence that you can make the small angle approximation.
{ In the small angle approximation, sin▯ ▯ tan▯ ▯ ▯;cos▯ ▯ 1.
{ In the small angle approximation, the vertical motion will be much
smaller than the horizontal motion, so assume there is no acceler-
ation in the vertical direction and just consider the motion in the
HORIZONTAL \x" direction.
(This can be proved by using Taylor expansions for cos▯ and sin▯.
The displacement in the x direction is: ▯x = l sin▯ ▯ l▯. The
displacement in the y direction is: ▯y = l(1 ▯ cos▯) ▯ l(1 ▯ (1 ▯
▯ =2)) = l▯ =2. Since ▯ is small, ▯ << ▯ so ▯y << ▯x.)
▯ Because this system is fairly simple and has a lot of symmetry, we can
▯gure out the normal modes by using some intuition. The idea is to
think up scenarios where the 2 pendulum bobs would oscillate at the
same frequency (so have the same period). There are 2 possibilities for
this system. They are:
▯ Symmetric case: (Fig. a above): Pendulums A and B are started at
equal angular displacement and released. Initial condition is x0A = x 0B ;
x_0A = 0, _0B = 0.
2 In this case, the spring always remains unstretched and so their is no
net spring force on the pendulum bobs. The only restoring force acting
on the bobs is gravity.
To ▯gure out the frequency of the system, we realize that the pendulums
do not a▯ect each other, and hence, the frequency will be the natural
frequency of a single pendulum in isolation. Solving F = ma in the x
and y directions for mass A gives:
Fnet;x ▯T sin▯ =Amx A
F = T cos▯ ▯ mg = 0
net;y A
(where we have used the fact that in the small angle approximation,
ay▯ 0. Using the small angle approximation for cos▯, setting sin▯ =
xA=l and solving for T in the second equation and plugging into the
▯rst gives us:
g
A xA= 0
l
(NOTICE: there is no x B in this equation). This is our familiar SHM
equation whose natural frequency is given by:
g
!sym = ▯ ! 0
l
Solving F net= ma for mass B would have given the same result. So
both A and B are oscillating at the same frequency !0.
▯ Antisymmetric case: (Fig b above) Pendulums A and B are started
at equal angular displacements in opposite directions and released. Ini-
tial condition is0A = ▯x 0B; _0A = 0, _0B = 0.
In this case, the spring is stretched a distancA 2x = Bxso there is a
spring force on each pendulum bob.
The net force on mass A in the x and y directions are:
Fnet;x= ▯T sin▯ A 2kx =Amx A
3 Fnet;y T cosA ▯ mg = 0
Again using the small angle approximation, we ▯nd:
▯ ▯
2k g
A+ + xA= 0
m l
(NOTICE: there is no B in this equation). This is also the equation
for SHM with a natural frequency of:
2k g
!2 = +
asym m l
If we solved for the motion Bfwe would ▯nd the same frequency.
Using our de▯nition of ! above, we can write the antisymmetric fre-
0
quency as:
r
2 2k
! asym= !0+ ▯ !0
m
Notice that the increased restoring force due to the spring INCREASED
the frequency of this vibration compared to the symmetric case above.
▯ Important point: once the motion has begun, neither motion will
change (in the absence of damping).
▯ S

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