Lecture 3: Damped SHM
1. Damping
The SHM equation mẍ = −kx gives a solution x(t) = Asin(ωt + φ) which oscillates for ever. This is
unrealistic: although some oscillations do go on for a long time, all slow down to some extent (and some
quite a lot). There is damping due to friction or friction-like forces which dissipate energy, and are hence
called dissipative forces. (‘Friction-like’ covers things like resistance in an LC circuit.)
The damping force depends on the speed v. It doesn’t act when the object is stationary. It acts against
the direction of the velocity to slow it down - so it is negative when v is positive and positive when v is
negative. It is presumably an increasing function of v - high speed means higher resistance.
Assume the resistive force is linear in v: F = −bv This satisfies all the above (zero at v = 0, increasing
with |v|, always opposing the motion.) And in some cases it is exactly true. In some cases it is true to a
good approximation. (In some cases it isn’t - for air resistance the force is actually much better modeled
by −bv − cv 2 .) With that assumption the equation of motion is mẍ = −bẋ − kx which can be more neatly
written as
mẍ + bẋ + kx = 0
(1)
2. Solving the Equation of motion
From observation, the motion is oscillatory but decays as time goes by. Decay is naturally described by
an exponential form: the fractional loss per time cycle always the same. So let’s try
γ
γ
x(t) = Ae− 2 t eiωt = Ae(iω− 2 )t
(The 2 is put in now to simplify later algebra.) Differentiating this just brings down a factor iω −
γ
substituting Equaion (2) in Equation (1) gives (after canceling Ae(iω− 2 )t )
m(iω −
(2)
γ
2.
So
γ 2
γ
) + b(iω − ) + k = 0
2
2
multiplying out that first bracket, and collecting real and imaginary parts, gives
−mω 2 + m
γ2
bγ
−
+k
4
2
+ i(−mωγ + bω) = 0
Both the real part and the imaginary part of this equation must be true. The imaginary part gives
b = mγ and thus
b
γ=
(3)
m
Substituting back for γ in the real part, the second and third expressions are almost the same, except one
has a 2 and one has a 4 so they don’t exactly cancel. We are left with
−mω 2 − m
γ2
+k =0
4
This gives the expression for the angular
p frequency ω which can be written in various ways. It’s useful to
bring in the undamped frequency ω0 = k/m.
ω 2 = ω02 −
γ2
k
b2
=
−
4
m 4m2
(4)
So: we have found a form of the equation, which validates our guess. The system oscillates at a frequency
ω which isn’t the same as the undamped frequency ω0 , though if the damping is small the difference is
negligible thanks to subtraction in quadrature. The damping term e−γt/2 = e−bt/2m causes the oscillations
to decay with time, and the bigger the damping b the faster the decay.
If you don’t like the complex form you can use the trig equivalent and put Ae−γt/2 sin(ωt) in Equation
(1). This gives an equation with a sin(ωt) and a cos(ωt) term, and making both of these vanish leads to the
same two equations as the imaginary and real parts did in the derivation above.
Lecture 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Damped SHM
3. Defining Q
γ has units of seconds−1 , like ω, and it makes sense to combine them into a quantity Q = ω0 /γ
Q (for Quality) expresses the rate of loss per cycle. A high Q means the oscillation persists for many
cycles, a low Q means it dies away rapidly.
The plot shows the pure sine wave, and decay for Q = 100, 10, 5 and 1. The lower plot is a close-up of
the first 1.5 cycles.
ω0
In terms of Q the damping term is e− 2Q t . As such that’s not very enlightening. But we can write the
time t in terms of the number of cycles n , which is also dimensionless: t = nT = 2πn/ω . Then the damping
is e−2πnω0 /2Qω ≈ e−nπ/Q . So the amplitude just falls by e−π/Q per cycle.
The difference between ω and ω0 is small (negligible?) unless Q is pathologically small. This can be
seen from the lower figure, looking not at the height of the peaks but at where the oscillations cross the axis.
The Q = 1 intersections are not quite thep
same, but any difference between
p the others is hard to see.
The frequency can be written ω = ω0 1 − 1/(4Q2) and the factor 1 − 1/(4Q2 ) is shown as a function
of Q. This boring plot confirms that the difference between the undamped ω0 and the actual ω negligible
for any reasonably large Q
4. Energy
The total energy of an oscillating spring is 12 kA2 . The energy (taken at the points where v = 0 and
all the energy is potential) falls like e−γt - gets rid of that factor of 2: γ is the decay rate for the energy.
Alternatively, the loss can be reckoned as a factor e−2π/Q per cycle.
5. Critical Damping
For high Q - light damping - the oscillation barely decays, for low Q - heavy damping - each successive
oscillation is a shadow of its predecessor. When the damping gets heavy enough - Q = 1/2- the expression
for ω 2 goes negative
This is handled by the maths. ω 2 goes negative, so ω is imaginary - say ω = ±iβ and there are solutions
f (t) = Ae(−γ/2+β)t + Be(−γ/2−β)t
p
If you look at the equation for β = γ 2 /4 − ω02 you can see that it is never bigger than γ/2 so both
exponentials are always falling. The boundary - ω = 0, Q = 1/2, γ/2 = ω0 is termed critical damping. Above
this critical value there are no oscillations. There are times when this is a good thing - e.g. in galvanometers.
A critically damped system is ideal in that it goes to zero as fast as possible without overshoot.
–2–