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Class Notes
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Canada
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University of Toronto St. George
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Physics
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PHY354H1
(28)

Erich Poppitz
(25)

Lecture

Department

Physics

Course Code

PHY354H1

Professor

Erich Poppitz

Description

Lecture 9: Damping + Restoring Forces
▯ relevant reading: Morin 4.3
▯ dynamics when you have restoring forces and damping
Damped Oscillations
▯ We know that all real oscillators experience some sort of dissipative
damping, and that if there is nothing driving the motion, the oscilla-
tions will eventually stop.
▯ Our simplest model for a resistive force like friction or air drag is
F = ▯bx_;
drag
where b is a constant and the minus sign indicates the force is in the
direction opposite to the motion.
▯ The total net force will be the restoring force plus the drag force. For
example, if the restoring force is due to a spring (so Hooke’s law), then
the net force is
F net= ▯kx ▯ bv
and Newton’s law tells us that
mx = ▯kx ▯ b_
or
b k
+ x_ + x = 0 (1)
m m
▯ This is the ODE we need to solve for damped oscillation problems.
Notice it is of the form = F(x;x_) and therefore cannot be solved
using separation of variables.
▯ We are now going to rename the parameters in front _ and x to make
2
some of the algebra simpler later on. Let 2
= b=m and !0= k=m.
Then the equation becomes:
1 2
+ 2
x_ + !0x = 0;
▯ The 2 in the above equation will make sense later. Notice that ! is0
the frequency of oscillations for a mass on a spring when there is NO
damping. We will call this the \natural frequency" of the system.
▯ Since this is a HOMOGENEOUS, linear, 2nd order ODE, and we know
we need x such that a combination of x;x _; and has to give 0, the
only function that can do that is an exponential. So just like in our
previous undamped examples, we will suppose we have solutions of the
▯t
form x = Ae . Taking the ▯rst and second derivatives and plugging
them into the equation gives:
2 ▯t ▯t 2 ▯t
▯ Ae + 2
▯Ae + ! 0e = 0
Dividing out the common Ae ▯tfrom each term gives:
2 2
▯ + 2
▯ + ! = 0 (2)
▯ This equation is known as the \characteristic equation" for this
problem. Just like in last week’s lecture (oscillations with no damping),
it allows us to solve for the ▯ in the solution but not the A. Solving
this quadratic equation gives:
q
▯2
▯ (2
) ▯ 4! 0 q
▯ = = ▯
▯
▯ ! 0
2
(notice how the "2" in our de▯nition of b=m = 2
made the algebra
simpler).
▯ Just like last week, we get 2 possible solutions (one with the plus sign,
one with the minus). Again, this is good because the mathematicians
tell us that 2nd order linear homogeneous ODEs have 2 linearly inde-
pendent solutions.
▯ To write the general solution we write a linear combination of the 2
solutions:
2 ▯ ▯ ▯ ▯
p 2 2 p 2 2
▯
+
▯!0 t ▯
▯
▯!0 t
x(t) = Ae + Be
where A and B are constants determined by the initial conditions.
▯
t
▯ I can simplify the equation q little. If I pull out the e from both
2 2
expressions and de▯ne
▯
▯ ! ,0I get:
h i
x(t) = e▯
t Ae
t + Be ▯
t
▯ Now,
is real so the exponential in front is purely real and cannot
represent oscillations.
may or may not be real depending on the sce-
nario so this will determine whether we get oscillations. Let’s consider
all the possible cases for
:
1.
< ! :0In this case,
is imaginary. We can then write:
q q q
2 2 2 2 2 2
=
▯ ! =0 (▯1)(! 0
) = i (! ▯ 0 )
q
▯ De▯ne !~ ▯ !0▯
. Plugging this into our solution gives:
▯
th ~t ▯i~t
x(t) = e Ae + Be
▯ We see that the term in square brackets represents oscillations
with angular frequency !~. However, now the oscillations are mul-
▯
t
tiplied by a negative exponential e . As time increases, this
term will decrease exponentially. Since it multiplies the oscilla-
tory term in brackets, the amplitude of the oscillations must also
decay exponentially.
▯ Remember that we can write the complex exponential term in a
variety of ways. For example, another way to write the above
▯
t
solution is: x(t) = Ce cos(~t + ▯):
▯ Run damped.py with omega0=5, gamma=1. This shows the nat-
ural undamped oscillation, the envelope and the damped solution.
▯ This type of damped oscillation is known as being \Underdamped".
The \under" is because the damping force is not strong enough
to overpower the restoring force and so there are some oscillations.
Essentially there is time to get in a few good oscillations before
the damping has a signi▯cant e▯ect.
3 ▯ Some useful jargon: Logarithmic decrement: After every pe-
riod T = 2▯=! ~, the ampltiude
x(t +

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