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PHY354H1
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Erich Poppitz
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Physics

PHY354H1

Erich Poppitz

Winter

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PHY254 Lecture 8
Oscillations in Phase Space and Dynamics
▯ Readings: For Oscillation dynamics: Morin 4.2
▯ Simple Harmonic Motion Examples
Simple Harmonic Motion
▯ Last time: We studied how to solve linear second order homogeneous
ODEs.
▯ Standard example: Mass on a spring with no damping or driving forces:
m + kx = 0 (1)
▯t
▯ tried a solution of the form xwhere A and ▯ are unknowns
▯ found: s
k
▯ = ▯ ▯
m
▯ we learned nothing about A, but we now have 2 possible solutions for
▯. ODE theory says that a linear homogeous 2nd order ODE will have
2 linearly independent solutions. We’ve found both of these. The ▯nal
solution is the superposition of these 2 solutions. So:
h▯q k▯ i h ▯q k▯ i
x(t) = Aexp ▯ m t + B exp ▯ ▯ m t .
for some A and B are constants which can be determined by the initial
conditions.
▯ For our speci▯c example, ▯k=m < 0. So we could right the exponential
as: ▯▯q p ▯ ▯ ▯ q ▯
exp k=m ▯1 t = exp it k=m
q
Notice thatk=m appears in the exponential where the angular fre-
quency should be. That makes it the angular frequency. Luckily for
this example we know what the angular frequency should be (just from
q
preqious studies of springs) and it isk=m. So we can write
! = k=m and plug it into the solution.
1 ▯ There are a bunch of ways you can write the solution that are all
equivalent:
x(t) = Ae i!t+ Be ▯i!t
x(t) = C cos!t + D sin!t
x(t) = E cos(!t + ▯ ) 1
x(t) = F sin(!t + ▯ ) 2
i!t+3
x(t) = Re(Ge )
etc. The coe▯cients are determined by the initial conditions. Notice
there are always 2 unknown constants in the above expressions.
▯ This type of motion that is purely sinusoidal is called \Simple Har-
monic Motion" (SHM for short).
▯ Example: Using initial conditions to solve for constant: A
particle on a spring is initially displaced a distance 0 from the origin
and has an initial velocity v . If we write the solution as
0
x(t) = Ae i!t+ Be ▯i!t
▯nd the constants A and B. (You may have noticed that this form
involves complex exponentials. In nothing we have done above did we
say we only need the real component of the solution. But we know the
solution must be real. The only way this can be is if the constants A
and B themselves are complex and related in a speci▯c way. Lets solve
for them and see what we get.)
Initial conditions: At t = 0, x = x0and v = v .0
▯ Answer: Lets write out the position and velocity for all time:
i!t ▯i!t
x(t) = Ae + Be
_(t) = i!Ae i!t▯ i!Be ▯i!t
Plugging in the initial conditions we ▯nd:
x 0 A + B
2 v0= i!A ▯ i!B
Solving for A in the ▯rst equation: A = x ▯ B and plugging it into
0
the second equation gives:
v0= i!x 0 2i!B
Solving for B gives:
v0▯ i!x 0
B = ▯2i!
Multiplying by i=i in order to make the denominator real gives us:
(v ▯ i!x ) i 1 ▯ v ▯
B = 0 0 = x0+ 0i
▯2i! i 2 !
Plugging this into the equation for A gives:
▯ ▯
A = x ▯ B = 1 x ▯ v0 i
0 2 0 !
We have now solved for the constants of integration and found them to
be complex. Although they are complex, they are related in a special
i!t ▯i!t
way. Because e and e are complex conjugates, so are the con-
stants. This is the only way to make the solution for x real.
Our ▯nal solution is:
1 ▯ v0 ▯ 1▯ v0▯
x(t) = x0▯ i e i!+ x0+ i e▯i!t (2)
2 ! 2 !
The answer we get above should be identical to what we would get
if we had chosen any of the other forms for the solution above (e.g.
x(t) = C cos!t + D sin!t). Lets verify that by solving the exact same
problem with this representation and showing we get the same answer:
▯ Example: Using initial conditions to solve for constants: A
particle on a spring is initially displaced a dista0ce x from the origin
and has an initial velocity0v . If we write the solution as
x(t) = C cos!t + D sin!t
▯nd the constants C and D. (Notice this is the exact same problem as
the previous one).
3 ▯ Answer: Writing the position and velocity:
x(t) = C cos!t + D sin!t
x_(t) = ▯!C sin!t + !D cos!t
Plugging in the initial conditions we ▯nd:
x0= C
v = !D
0
So C = x 0nd D = v =0. The solution is
x(t) = x0cos!t + (v0=!)sin!t (3)
Is equation (3) the same answer as equation (2)? Check by using
DeMoivre’s theorem to write equation (2) in terms of sines and cosines:
1 ▯ v ▯ 1 ▯ v ▯
x(t) = x0▯ 0i ei!t+ x 0 0i e ▯i!t
2 ! 2 !
1 ▯ v0 ▯ 1 ▯ v0 ▯
= x0▯ i (cos!t + isin!t) + x0+ i (cos!t ▯ isin!t)
2 ! 2 !
= x 0os!t + (v 0!)sin!t
So yes they are the same. Notice that because the exponentials were
complex conjugates and the constants in front of the exponentials were
complex conjugates, all of the imaginary components cancel out of the
▯nal equation.
▯ In the previous lecture, we learned that the phase space curves of x =
Acos(!t+▯) are ellipses. Thus, simple harmonic motion in phase space
presents elliptical curves.
4 Studying the Simple Pendulum
Example: Pendulum \down": Consider a pendulum

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