Oscillations
Vern Lindberg
June 10, 2010
You have discussed oscillations in Vibs and Waves: we will therefore touch lightly on
Chapter 3, mainly trying to refresh your memory and extend the concepts.
1
Undamped Simple Harmonic Motion
If a mass m is attached to an ideal spring of constant k, and the spring force is the only
net force, then one dimensional simple harmonic motion (SHM) results. The equation of
motion is
k
ẍ = − x
(1)
m
The solution can be written is several forms.
x = A0 sin(ωt + φ0 )
(2)
= A1 cos ωt + A2 sin ωt
(3)
= A3 exp(iωt) + A4 exp(−iωt)
(4)
= A5 exp(i(ωt + φ5 ))
(5)
The angular frequency, ω = 2πf , is independent of amplitude and is
r
k
ω=
m
(6)
Since this is a second order differential equation, all solutions have two integration constants (such as amplitude and phase, A0 and φ0 , or two amplitudes A1 and A2 ) that are
determined by the initial conditions.
Picking Eq. 2, the first solution, If at t = 0 we have x = x0 and v = v0 then
x(0) = A sin φ0 = x0
(7)
ẋ(0) = ωA cos φ0 = v0
(8)
1
Solving these we get
ωx0
v
r0
tan φ0 =
x20 +
A =
(9)
v02
ω2
(10)
We will choose the positive square root value for amplitude, but there remains an ambiguity
in the inverse tangent that is resolved by deciding what quadrant for φ0 is needed to
produce the proper signs for initial position and velocity. We will spend some time on this
in class.
Example: Suppose the spring constant is 18 N/m and the mass is 2 kg. Then the angular
frequency is ω = 3 rad/s. If x0 = 1.675 cm and v0 = −14.13 cm/s, then the amplitude
is 5.0 cm and the inverse tangent gives phase angles of either -0.34 or 2.80 radians.
With a bit of thought you can decide that the correct phase is 2.80 radians.
Example Using Eq. 3, find the values of A1 , A2 Use the same values as in the previous
example.
The potential energy associated with this simple model SHM is
1
V (x) = kx2
(11)
2
This is a parabolic potential energy, with the oscillations occurring around the minimum
in the potential energy. The spring force is a restoring force, linear in displacement.
1.1
The Morse Potential for Diatomic Molecules: Approximations
Any potential energy that has a minimum can have oscillations around that minimum.
A natural question is to ask whether the oscillations are simple harmonic. If the system
oscillates about x = x0 with force F = −k(x − x0 ) and potential energy V = 21 k(x − x0 )2
then the motion is SHM for any amplitude. What if we don’t have forces of that form?
For small oscillations about the minimum we can make suitable approximations to see if
the small amplitude motion is SHM.
Taylor Series and Approximations A vital skill in physics is making approximations.
The Taylor series is a polynomial expansion of a function about some point. If we
expand f (x) about x = a, the Taylor series is
∂f 1 ∂ 2 f f (x) = f (a) +
(12)
(x − a) +
(x − a)2 + · · ·
∂x x=a
2! ∂x2 x=a
The terms are called the zeroeth order, first order, second order, . . . . We frequently
want to find the first higher order term (first and above) that is non-zero.
2
Consider the Morse potential energy, a model of the potential energy for a vibrating diatomic molecule like O2 .
x − x0 2
V (x) = V0 1 − exp(−
) − V0
(13)
δ
First we find the location of the minimum by solving dV /dx = 0 and find that the minimum
occurs at x = x0 . This is done as an example in Chapter 2 of the text. At this minimum
we find V = −V0 .
Computational Hint Make appropriate substitutions of variables to make the calculus
easier. For the Morse potential define z = (x−x0 )/δ. Then V = V0 (1−exp −z)2 −V0 ,
and we can do derivatives as dV /dx = dV /dz dz/dx.
For small oscillations we expand the exponential in a Taylor series,
V
= V0 [1 − (1 − z + z 2 /2 − z 3 /6 + · · · )]2 − V0
= V0 [z − z 2 /2 + z 3 /6 + · · · ]2 − V0
= V0 [z 2 − z 3 + 7z 4 /12 + · · · ] − V0
x − x0 2
2
= V0 z − V0 = V0
− V0
δ
(14)
This is parabolic and so we expect SHM. The restoring force for small oscillations is linear,
dV
2V0
F (x) = −
= − 2 (x − x0 )
(15)
dx
δ
Thus for small oscillation we expect SHM with an angular frequency of
r
2V0
ω=
(16)
mδ 2
Example The following system is set up on a horizontal, frictionless table. A small block
of mass m is attached to two identical springs of constant k. When the springs are
unstretched they have identical lengths L0 . The equilibrium position is a straight
line consisting of a wall, a spring, the block, the other spring, and a second wall. The
block is pulled down a small distance y L0 perpendicular to the equilibrium line.
Determine whether or not the subsequent motion is simple harmonic.
1.2
Angular SHM
Consider a simple pendulum of length ` with point mass m attached at one end, in a
uniform gravitational field ~g . The motion is along an arc of a circle, so it is natural to use
3
an angular variable θ for the displacement. Using the angular form of Newton’s Second
Law, in scalar form since the angular motion is along a fixed direction,
N
= I θ̈
(17)
2
−mg` sin θ = (m` )θ̈
(18)
Providing the angle (in radians) is small, we can use the Taylor Series expansion sin θ =
θ − θ3 /3! + · · · and keep it to first order to get the equation of motion.
g
θ̈ + θ = 0
`
(19)
This is the same form as the SHM
p equation of motion, Equation 1, so the solution is SHM
with angular frequency ω = g/`. We will solve the general equation of motion (large
angles) in Mechanics 2 using Elliptic Integrals.
1.3
Energy and SHM
For ideal SHM the energy is a constant of the motion and is
1
1
E = mẋ2 + kx2
2
2
2
(20)
Damped Simple Harmonic Motion—Linear Damping
Suppose we add a linear damping term to the equation of motion to get
ẍ = −
c
k
ẋ − x
m
m
(21)
c
It is convenient to define a damping factor γ = 2m
so that the equation of motion becomes
ẍ + 2γ ẋ + ω02 = 0
(22)
Next let’s introduce a differential operator, D = d/dt. The operator has no independent
meaning, but once it is applied to a function its meaning is clear. The equation of motion
can be written
2
D + 2γD + ω02 x = 0
(23)
q
q
D + γ − γ 2 − ω02 D + γ + γ 2 − ω02 x = 0
(24)
There are two solutions, one for the first bracket operating on x equalling 0 and one for
the second bracket operating on x being zero.
4
These are first order differential equations and are easy to solve resulting in a general
solution
x(t) = A1 exp [−(γ − q)t] + A2 exp [−(γ + q)t]
(25)
p
2
where q = γ 2 − ω0 .
Case 1 When γ 2 − ω02 > 0, q is real and positive and we have the overdamped case,
where a displaced mass slowly returns to equilibrium with no oscillations. Note that
(γ − q) > 0.
Case 3 When γ 2 − ω02 < 0, q is imaginary and we have the underdamped case where the
mass oscillates about equilibrium with a decreasing amplitude. Defining
q
ωd = ω02 − γ 2
(26)
and doing some work (shown in the text) we can write the solution can as
x(t) = e−γt A cos(ωd t + θ0 )
(27)
where A and θ0 are determined by initial conditions.
Case 2 When γ 2 − ω02 = 0 we have critical damping. The mass quickly returns to equilibrium without overshooting.
If we start with Equation 25, we run into a problem: we have only one arbitrary
integration constant. To find a solution with two integration constants we must
return to the differential equation, Equation 24, that becomes
[D + γ] [D + γ] x = 0
(28)
This is a second order differential equation and we need two arbitrary constants.
Here is the trick needed to get the general solution. Define u = (D + γ)x. Then
(D + γ)u = 0 and this has the solution u = Ae−γt . The solution for x is found from
Ae−γt = (D + γ)x
A = eγt (D + γ)x = D(xeγt )
xeγt = At + B
x = Ate−γt + Be−γt
(29)
Example Suppose we have a mass of 3.0 kg attached to a spring of constant 12.0 N/m,
and with initial conditions that at t = 0, position is 1.0 m and velocity is 0.0 m/s.
(a) For the underdamped case with c = 4.0 kg/s find the values of the integration
constants. [A = 1.06 m, θ0 = -0.34 rad] (b) For the critically damped case with
c = 12 kg/s find the values of the integration constants. [A = 2.0 m/s, B = 1.0
m] (c) For the overdamped case of c = 20 kg/s, find the values of the integration
constants. [A1 = 1.125 m, A2 = −0.125 m]
5
2.1
Energy and Linear Damping
The instantaneous energy is still E = mẋ2 /2 + kx2 /2 so we can look at how this changes
with time.
dE
= mẋẍ + k ẋx = (mẍ + kx)ẋ
(30)
dt
But from Equation 21, mẍ + kx = −cẋ, hence
dE
= −cẋ2
dt
2.2
(31)
Quality Factor for Damped SHM
For the underdamped case we can characterize the system by its Q-factor,
E Q = 2π ∆E (32)
where E is the energy stored in the oscillator (think in terms of the average energy) and
∆E is the energy lost in one period of the oscillation.
Q is calculated usually for weakly damped systems where the energy does not change
quickly. In this case the energy at time t can be written
1
1
E = kA2 e−2γt = mω02 A2 e−2γt
2
2
(33)
Using Equation 27 for x we differentiate to get
ẋ = −Ae−γt (γ cos(ωd t + θ0 ) + ωd sin(ωd t + θ0 ))
(34)
Then, changing variables to z = ωd t + θ0
Z
Td
−cẋ2 dt
0
Z
cA2 2π+θ0 −2γt 2
= −
e
γ cos2 z + 2γωd cos z sin z + ωd2 sin2 z dz
ωd θ0
∆E =
(35)
(36)
But for weakly decaying systems the exponential changes very little during the integration,
so we remove it from the integral. The integration limits can start anywhere as long as
they extend over 2π. So
Z
cA2 −2γt 2π 2
∆E = −
e
γ cos2 z + 2γωd cos z sin z + ωd2 sin2 z dz
(37)
ωd
0
6
The integrals for cos2 and sin2 each yield π, and the cross term becomes 0, giving
∆E = −
Then
πcA2 −2γt 2
πcA2 −2γt 2
e
γ + ωd2 = −
e
ω0
ωd
ωd
1
mω 2 A2 e−2γt
ωd
=
Q = 2π 2πcA20
2
−2γt
2γ
ω0
ωd e
(38)
(39)
Table 3.4.1 in the text lists some Q’s for underdamped systems. The earth (responding to
earthquakes) has Q = 250 to 1400 while a piano string has Q = 3000.
2.3
An Example
Suppose a mass of 500 kg is attached to an ideal spring of force constant 2000 N/m and a
linear damping shock absorber of constant 20 N s/m. Find
(a) The natural (undamped) angular frequency and frequency in Hz.
(b) The frequency with damping applied.
(c) The Q of the system
(d) The time for the amplitude to reduce to 1/10 of its original value.
(e) The energy when the amplitude has reduced to 1/10 its original value.
We have m = 500 kg, k = 2000 N/m, and c = 20 N s/m.
p
(a) Use ω0 = k/m = 2.00000 rad/s. This gives a frequency of f = ω0 /2π = 0.318310 Hz
(and a period of 3.14159 s).
p
(b) Compute γ = c/2m = 0.02 rad/s. Then ωd = ω02 − γ 2 = 1.99990 rad/s.
(c) The quality factor is Q = ωd /2γ = 49.997.
(d) The envelope of the oscillation varies as A(t) = Ae−γt , so solve 0.1 = e−0.02t to get
t = 115.13 s. (37 oscillations during this time.)
(e) Since E ∝ A2 , when the amplitude is reduced to 1/10 of its original value, the energy
is reduced to 1/100 of its original value.
3
Phase Space
Plots of position versus time and velocity versus time are useful, but often a plot of velocity
(or momentum) versus position is drawn. A plot such as this is called a phase plot.
7
For undamped SHM, x = A sin(ω0 t + φ0 ) and ẋ = Aω0 cos(ω0 t + φ0 ). We combine these
so as to eliminate time and get the parametric equation
x2
ẋ2
+
=1
A2 A2 ω02
(40)
which is the equation of an ellipse. Figure 3.5.1 shows several trajectories in phase space
corresponding to different initial conditions. All trajectories in this situation form closed
paths.
For underdamped oscillations the phase plot spirals in toward a final value of v = 0.
The text does some very clever changes of variables to make the phase plot more obvious:
I will just demonstrate using Easy JAVA Simulation, and extend to the case of critical
damping and over damping.
Non linear oscillators can lead to chaotic motion, a topic left for other courses.
4
Forced Damped Harmonic Motion
Consider the damped harmonic oscillator subject to a periodic driving force F0 cos ωt. The
equation of motion is
mẍ = −kx − cẋ + F0 cos ωt
(41)
First consider undamped motion and guess the solution to be x = A cos(ωt − φ) with two
variables, A, φ to be determined.
The equation of motion becomes
− mAω 2 cos(ωt − φ) + kA cos(ωt − φ) = F0 cos ωt
(42)
This only works for the two values φ = 0, π, implying
A =
=
F0 /m
ω02 − ω 2
F0 /m
ω 2 − ω02
φ=0
Low frequency
(43)
φ=π
High frequency
(44)
This gives a resonance at ω = ω0 .
When we add damping and the velocity term we have difficulty when starting with a
cosine solution. Instead we assume a driving force as F0 eiωt and consider the real part of
8
the complex quantity to describe the physical situation. We guess a solution x = Aeiωt−φ
so that the equation of motion becomes
− mω 2 Aeiωt−φ + kAeiωt−φ + icωAeiωt−φ = F0 eiωt
2
−mω A + kA + icωA = F0 e
iφ
= F0 (cos φ + i sin φ)
(45)
(46)
The real parts must be equal, and the imaginary parts must be equal giving
A(k − mω 2 ) = F0 cos φ
(47)
c ω A = F0 sin φ
(48)
These can be solved simultaneously and give
tan φ =
A =
2γω
− ω2
ω02
F0 /m
p
2
(ω0 − ω 2 )2 + 4γ 2 ω 2
(49)
(50)
Further details should have been discussed in Vibrations and Waves. You will probably
see resonance in an electrical circuit when you take Electronic Measurements.
When the resonance is fairly sharp, meaning relatively small damping, the resonant frequency is very close to the natural frequency. For these cases we define the sharpness of
the resonance with another Q factor—same symbol and definition as before, but a different meaning. If the resonance has an amplitude Amax at resonance, ω0 , √
then there
are frequencies above and below the resonant frequency at which A = Amax / 2. These
two frequencies are called the half-energy (or half-power) frequencies, and the difference in
frequencies is
c
∆ω = 2γ =
(51)
m
and
ω0
ω0
Qresonance =
≡
(52)
∆ω
2γ
9