PHY141 Lectures 6,7 notes

Alice Quillen

September 22, 2022

Contents
1 Harmonic Motion 1
1.1 The harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Sine and Cosine coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Amplitude and phase coefficients . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 With complex exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Trajectories of the Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . 5
1.6 The pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Some Force diagrams 8

3 Damped and driven harmonic motion 9

3.1 Damped harmonic motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
3.2 Driven harmonic motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.3 Driven and damped harmonic motion . . . . . . . . . . . . . . . . . . . . . 13

4 Springs in parallel and in series 14

4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5 Atomic forces in solids, stress and strain 16

5.1 Stress and strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
5.2 Deviations from linear elasticity . . . . . . . . . . . . . . . . . . . . . . . . . 19

6 Summary 19

7 Props 20

1 Harmonic Motion
We focus on a point mass upon which we exert a force. Previously we looked at a constant
force, like gravitational acceleration on the surface of a planet, and inverse square law

1
forces (gravity or electric). A very simple force law that depends linearly on position in
1 dimension
F (y) = −k(y − L)
where positive k is a spring constant in units of N/m and the spring rest length is L.
Spring forces for actual springs are not exactly linear and the force might also depend on
velocity as well as position.
If we shift the coordinate system so that x = y − L is a displacement from rest, then
the force law is even simpler
F (x) = −kx
This is known as Hooke’s law. The force is applied in the direction opposite to the spring
displacement.

Figure 1: A spring force is linearly dependent on position. Here the block is assumed to
be resting on a frictionless surface.

Many materials respond approximately linearly, with force proportional to displace-

ment. It is challenging to design a mechanism with a constant force that is indepen-
dent of displacement. For a clever mechanism that can be 3D printed see https://www.
thingiverse.com/thing:4624094.

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1.1 The harmonic oscillator
Consider a point mass of mass m that is connected to a massless spring with fixed endpoint.
The value of x gives the position of the mass. We ignore gravity. The equation of motion
using F = ma is

d2 x
m = −kx
dt2
d2 x k
=− x
dt2 m
The general solution can be written in different ways

x(t) = A cos(ωt) + B sin(ωt)

= Ceiωt + De−iωt
= a cos(ωt + φ0 )

with angular frequency r

k
ω=
.
m
Depending upon which form you chose, the solutions depend on coefficients A, B or am-
plitudes C, D or amplitude a and phase φ0 .

1.2 Sine and Cosine coefficients

Using the form
x(t) = A cos(ωt) + B sin(ωt)
we differentiate x(t) to find the velocity

dx
v(t) = = −Aω sin(ωt) + Bω cos(ωt)
dt
Using initial conditions x(t = 0) = x0 and v(t = 0) = v0 we find that

x0 = A v0 = Bω.

We solve for A, B in terms of x0 , v0 .

v0
A = x0 B=
ω
The solution at later times
v0
x(t) = x0 cos(ωt) + sin(ωt)
ω

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1.3 Amplitude and phase coefficients
Using the form
x(t) = a cos(ωt + φ0 )
we differentiate x(t) to find the velocity

v(t) = −aω sin(ωt + φ0 ).

At t = 0, the displacement and velocity

x0 = a cos φ0 v0 = −aω sin φ0

We solve for φ0 , a
r
v2
a= x20 + 02
ωv 
0
φ0 = atan2 − , x0
ω
Note the atan2 function gives an angle within [−π, π] or [0, 2π]. The arctan function alone
would only give an angle within [−π/2, π/2] or [0, π].
The solution at later times
r
v2   v
0

x(t) = x20 + 02 cos ωt + atan2 − , x0
ω ω

1.4 With complex exponentials

Using the form
x(t) = Ceiωt + De−iωt
we differentiate x(t) to find the velocity

v(t) = Ciωeiωt − Diωe−iωt

At t = 0 the coefficients
x0 = C + D v0 = iω(C − D)
We solve for the coefficients
1 v0 
C= x0 +
2 iω
1  v0 
D= x0 −
2 iω

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The solution at later times
1 v0  iωt 1  v0  −iωt
x(t) = x0 + e + x0 − e
2 iω 2 iω
It may be disturbing to describe a real system with complex numbers for displacement
and velocity. However it can be convenient to solve the problem in a complex form and
then assert that the actual solution is the real part of the complex one. Alternatively, by
requiring the initial conditions to be real, a real solution at later times is ensured.

1.5 Trajectories of the Harmonic oscillator

Trajectories are shown in different ways in Figure 2. We draw x(t), v(t) vs t and we draw
x(t) vs v(t) (which is known as phase space).
What is the period?
2π
P =
ω
The period is independent of oscillation amplitude.
Note that
x2 + v 2 /ω 2 = constant
and this is equivalent to conservation of energy. The trajectory is an ellipse in a v vs x
plot.

Figure 2: Trajectories for the harmonic oscillator. Displacement and velocity are sinusoidal
(on the left) whereas the trajectory is an ellipse in phase space (on the right).

This physical model is known as the harmonic oscillator and is a ubiquitous model.
If you can make a complicated problem look like a harmonic oscillator, then you can
(approximately) solve it.

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1.6 The pendulum
We consider a point mass of mass m that is attached to a massless string of length L. The
end of the string is held fixed. The point mass feels gravitational acceleration. The state
of the system is described with an angle θ.

Figure 3: The radial force component is balanced by tension on the string. The tangential
force component is F = mg sin θ.

We decompose the force from gravity (onto m) into radial and tangential components

F = −mgẑ = mg cos θr̂ − mg sin θθ̂

The radial component is balanced by the tension in the string leaving the tangential com-
ponent to accelerate the mass m.
Recall that in polar coordinates

a = (r̈ − rθ̇2 )r̂ + (2ṙθ̇ + rθ̈)θ̂

As ṙ = 0 and the radial component is balanced by tension we are left with

a = Lθ̈θ̂

(here radius is L). Using our tangential force component

mLθ̈ = −mg sin θ

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 With the small angle approximation

sin θ ≈ θ

and we find
g
θ̈ = − θ.
L
We can solve this equation of motion with
r 
g
θ(t) = θ0 cos t + φ0
L

with phase φ0 . The frequency of oscillation is

r
g
ω=
L

and the oscillation period is s

2π L
P = = 2π .
ω g
With small amplitudes of oscillation, the period is independent of amplitude and we
have equations the same as for a harmonic oscillator. However when the amplitude is large
the period depends on amplitude.
For a pendulum the period depends on the amplitude with higher amplitudes having
longer periods. Is it possible to design a pendulum that has period independent of ampli-
tude? Then answer is yes. See the cycloidal pendulum animation and The tautochrone
curve.

Figure 4: The cycloid pendulum.

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2 Some Force diagrams
Some force diagram examples to discuss in class.

Figure 5: We assume the system is in an equilibrium state. The tension force in the wire
T = T cos θx̂ + T sin θẑ. The z component of tension is balanced by gravity and the x
component is balanced by the spring force. Here +z is upward and +x is to the right.

Figure 6: The three blocks are on a frictionless surface. The force F = (m1 + m2 + m3 )a.
The acceleration of all the blocks is the same. We call the contact force f1,2 that between
blocks 1 and 2, and f2,3 the contact force between blocks 2 and 3. The contact forces are
exerted equally and oppositely between pairs of blocks.

All three blocks in Figure 6 have acceleration a. This means that

F = (m1 + m2 + m3 )a.
The leftmost block (m1 ) must have contact force
f1,2 = m1 a
on it because there are no other horizontal forces on it. The rightmost block (m3 ) has
m3 a = F − f2,3 = (m1 + m2 + m3 )a − f2,3 .

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This means that the contact force

f2,3 = (m1 + m2 )a,

which makes sense as f2,3 must account for acceleration of m1 + m2 . The middle block has

m2 a = f2,3 − f1,2 = (m1 + m2 )a − m1 a,

as expected.

Figure 7: Left: A spring network. If the network is in equilibrium, the sum of the force
vectors at each node is zero. Right: An illustration of a chain of transmission of stress
forces through contacts in a granular media. The force chain figure is by Gsrdzl - Own
work, CC BY-SA 3.0. and licensed under Creative Commons Attribution-Share Alike 3.0
Unported.

3 Damped and driven harmonic motion

3.1 Damped harmonic motion
Often there are velocity dependent forces as well position dependent forces. We modify
our spring model so the force is
F = −kx − bv. (1)

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This first term is a spring force. The second term is a velocity dependent force that depends
on positive coefficient b that would be caused by a dashpot. The equation of motion
d2 x b dx k
+ + x = 0. (2)
dt2 m dt m
It is useful to define
r
k
ω0 ≡
m
b
γ≡ (3)
m
with γ in units of inverse time. The equation of motion (equation 2 becomes)

ẍ + γ ẋ + ω02 x = 0. (4)

We assume a general solution

x(t) = Aeiωt
Here ω can be positive or negative and A can be complex.
What does it mean to assume a complex solution when x is a real distance? If the
equations are linear then we can solve for complex x and then take the real part.

Re eiωt = cos ωt.

If A is complex then you need to take that into account when you take the real part of
Aeiωt .
Insert our general solution into the equation of motion (eqn 4)

−ω 2 Aeiωt + γAiωeiωt + ω02 Aeiωt = 0

ω 2 − γiω − ω02 = 0.

We solve for the frequency ω using the quadratic equation

iγ 1
q
ω= ± 4ω02 − γ 2 . (5)
2 2
We have three cases for solutions
• γ < 2ω0 . This case is known as weakly damped. The argument inside the square root
is positive so the general solution is
h γt γt
i
x(t) = Re Ae− 2 eiω̃t + Be− 2 e−iω̃t

with
1
q
ω̃ = 4ω02 − γ 2 .
2

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 • γ > 2ω0 . This case is known as over-damped. The argument inside the square root
is negative so the general solution is
(γ+ν)t (γ−ν)t
h i
x(t) = Ae− 2 + Be− 2

with
1
q
ν= γ 2 − 4ω02
2
.
• γ = 2ω0 . This case is known as critically damped. The general solution is
γt γt
x(t) = Ae− 2 + Bte− 2 .

We can write t
γt −t
e− 2 = e damp

in terms of a damping timescale

2 2m
tdamp = = .
γ b
The amplitude of the weakly damped solution decays on this exponential timescale.

Figure 8: The different types of solutions for the damped harmonic oscillator. The red
lines show a highly damped harmonic oscillator. The blue lines show a weakly damped
harmonic oscillator. On the left we show x(t) vs t. On the right we show phase space;
dx/dt vs x.

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3.2 Driven harmonic motion
Consider a damped harmonic oscillator that driven by a sinusoidal force

Fdriving = ma0 sin ωt

that is pushing on the mass. Here a0 is in units of acceleration. We modify equation 2 to

include this driving force. The damped driven harmonic oscillator has equation of motion
d2 x dx
2
= −ω02 x − γ + a0 sin(ωt). (6)
dt dt
We start with the oscillator with some initial position and velocity. The solution will have
a transient response which decays and then will approach a steady state, which likely will
be sinusoidal and described with a particular amplitude, frequency and phase.
To make things simpler, we will ignore the transient response and we look at driven
harmonic motion without any damping. With b = γ = 0, the equation of motion is
d2 x
= −ω02 x + a0 sin(ωt). (7)
dt2
The steady state solution (ignoring transient response) must depend on the driving
frequency ω
x(t) = A cos(ωt) + B sin(ωt). (8)
We insert this into the equation of motion

−Aω 2 cos(ωt) − Bω 2 sin(ωt) = −Aω02 cos(ωt) − Bω02 sin(ωt) + a0 sin ωt

We get two equations

−Aω 2 = −Aω02
−Bω 2 = −Bω02 + a0

The first equation gives A = 0 (unless ω = ω0 ) and the second one gives
a0
B=
ω02 − ω 2
Units are ok! Inserting our solution for B into equation 8 we find a solution for the
undamped but driven harmonic oscillator
a0
x(t) = sin ωt. (9)
ω02 − ω2
This is a solution to the inhomogeneous ordinary differential equation 7. The general
solution would be a sum of homogeneous and inhomogeneous terms. (That means we can
add any solution of the non-driven harmonic oscillator to get another solution).

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 If ω = ω0 we have a problem as the amplitude is infinite! Had we used a damped
spring, the response would have been limited by the damping rate.
This is an example of resonant response, as there is a strong response near a resonant
frequency.
Notice the response is either in phase or with the opposite phase as the driving force.
The sign of the response depends on whether the driven frequency is larger or smaller than
the resonant one.
In the adiabatic limit (slow frequency driving) the solution is B = a0 /ω02 . In this limit
ω0 can be considered large and the spring constant is large. We balance kx against the
driving force (and neglected the inertial force that depends on mass) and this gives the
solution.
In the opposite limit (fast frequency driving) the amplitude is small – the forcing
averages to zero.
We have assumed an infinite time period for steady driving. There would be transients
associated with the onset of the driving force in a real system.
With both damping and forcing, the near resonant response is not infinite and the
phase varies continuously from −π to π as the driving frequency passes across the resonant
frequency.

3.3 Driven and damped harmonic motion

The driven and damped harmonic oscillator

d2 x dx F
+ 2ζω0 + ω02 x = sin(ωt) (10)
dt2 dt m
(with ζ = 2ωγ 0 = 2mω
b
0
of our damped oscillator; equations 1 and 2). The steady state
solution can be written in the form

x(t) = A cos(ωt + φ)

with amplitude A and phase φ being functions of driving force F , frequency ratio ω/ω0 and
damping parameter ζ. What do we mean by steady state? We mean one with constant
amplitude, not one that does not oscillate. Independent of its initial conditions, a driven,
damped harmonic oscillator will approach this solution. The amplitude of the steady state
solution satisfies
F0 1
A= q . (11)
mω (ω 2 −ω 2 )2
(2ω0 ζ)2 + 0 ω2
The amplitude of the steady state solution of equation 10 is shown in Figure 9 for different
values of damping parameter ζ. I wrote the damping term in terms of ζ so as to be
consistent with Figure 9.

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Figure 9: The damped driven harmonic oscillator. The x axis is in units of ω/ω0 . The
y axis shows the amplitude of the steady state sinusoidal solution. Here ω is the driving
frequency, ω0 is the resonant frequency and ζ is a damping parameter. The dotted grey
line goes through the maxima of each curve. The dashed grey lines show an envelope of
possible solutions. This figure is a modified version of Mplwp resonance zeta envelope.svg.
Axes have been more clearly labelled. The original file is licensed under Creative Commons
Attribution 3.0 Unported license.

The phase difference between driving frequency and harmonic oscillator satisfies
γω 2ω0 ζω
tan φ = − =− 2 . (12)
ω02 − ω 2 ω0 − ω 2

4 Springs in parallel and in series

4.1 Two springs in series
Consider two springs k1 and k2 in series (see Figure 10) with the seconnd spring connected
to a mass m. The first spring is stretched by x1 and the second spring by x2 . The total
displacement is x1 + x2 . What is the effective spring constant ks of both springs? Meaning
if we think of them as a single spring what would its spring constant be?
If the springs are massless and the bottom one is under a load or weight, then both
springs have the same force on them.

F = k1 x1 = k2 x2 = ks (x1 + x2 ).

Equivalently look at the point between the springs. The forces k1 x1 = k2 x2 because
equal and opposite forces apply from both springs at this point. The force exerted by the

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 1
Figure 10: Massless springs in series. The effective spring constant ks = k1−1 +k2−1
.

first spring on the second spring must be equal and opposite to that exerted by the second
spring on the first spring.
Look at the point touching the mass m. We know that k2 x2 is the force on m and this
must be equal to ks (x1 + x2 ) as this too must be the force on m.
Using k1 x1 = k2 x2 , we solve for x2
k1
x2 = x1 .
k2
Using ks (x1 + x2 ) = k1 x1 , we solve for the effective spring constant ks
k1 x1
ks =
x1 + x2
k1 x1
=
x1 + (k1 /k2 )x1
k1 k2
=
k1 + k2
1
= −1 . (13)
k1 + k2−1
There is an analogy for resistors in parallel in electric circuits.

4.2 Two springs in parallel

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 Figure 11: Massless springs in parallel. The effective spring constant ks = k1 + k2 .

Springs in parallel are more straight forward. The displacements from both springs are
the same. The forces add. F = (k1 + k2 )x = ks x. The effective spring constant for two
springs in parallel
ks = k1 + k2 . (14)
The spring constants add. There is an analogy for resistors in series.

4.3 Example
We start with a single spring with spring constant k. We consider a chain of these springs,
in series, that is 100 springs long. What is the spring constant of the chain?

1 k
kchain = P100 1
= .
i=1 k
100
We have a block that is comprised of 2000 such chains in parallel. What is the spring
constant of the 2000 chains?

k
kblock = kchain × 2000 = × 2000 = 20k.
100

5 Atomic forces in solids, stress and strain

Consider a metal with density ρ and a molar mass of Mmolar . The density is mass per unit
volume. The molar mass is the mass of a mole or Avogadro’s number of atoms. Avogadro’s
number is
atoms
NA = 6 × 1023 .
mol

16
Figure 12: Bond strengths are modeled with springs. A mass spring model approximates
the elastic properties of a solid. However, spring forces connect between point masses
so sensitivity to bond angle is neglected. This model is classical and neglects thermal
fluctuations. Here the inter atom spacing is l and the length of the cube is L. The number
of atoms in length L is NL = L/l. The number density n = NL3 /L3 = l−3 .

A mole of this metal has how much volume?

1 mass mol mol
ρ× = × =
Mmolar volume mass volume
The volume of a mole
Mmolar volume
Vmole = = .
ρ mol
The number density is the number of atoms per unit volume; n = N/V where N is a
number and V is a volume. For this metal, what is the number density of atoms, n?
NA 1
n= =ρ× × NA
Vmole Mmolar
mass mol atoms
= × ×
volume mass mol
atoms
=
volume
What is the typical length, l, between atoms in our metal? Consider a cubic volume
that is V = L3 with L a side of the cube. Then the number of atoms in the cube is
N = nL3 . We arrange the N atoms in a cubic lattice. Then number of atoms on a side of

17
Figure 13: Estimating interparticle force strength from the stretching of a wire. We count
the number of atoms in a single layer in the base. Each linear segment can be considered
a single long spring.

the cube is NL with N = NL3 and L = NL l. This means that

N = nV
NL3 = nL3
NL 3
 
=n
L
L 3
 
= n−1
NL
 
L
= n−1/3
NL

We compute l = L/NL which is the length per atom on a side and this we find is equal to
1
l = n− 3 . (15)

What is the typical length between atoms in our metal?

 1
− 13 Mmolar /ρ 3
l=n = . (16)
NA
To estimate the force strength between atoms we stretch a wire and measure the force
required to strength it by a particular length. The wire is modeled as Na single long
springs. Each long spring has NL atoms in it. Using our formulas for how spring constants
add in parallel and in series we can relate the total distance stretched to the amount the
distance between atoms is stretched. We can estimate the spring constant between two
atoms.

18
 Na , is the number of single long springs. We estimate this by dividing the area of the
wire by the inter atomic spacing l2 . NL is the number of atoms in a single long chain that
is the length of the wire. We estimate this by taking the length of the wire and dividing it
by l.

5.1 Stress and strain

Stress is force per unit area. However force is a vector and the direction that it is applied
(or the normal vector to the area) is also a vector. So stress is a tensor. Compressive
stress is exerted when the entire volume shrinks. Tensile stress is exerted when the volume
expands, for example when a wire is pulled. Sheer stress is when a cube of jello is pushed
on the top so that it tilts.
Strain is a deformation per unit length. It is related to a gradient of displacement. A
particle can be displaced w.r.t. to another particle in 3 directions, compared to the vector
between the two particles. So strain is also a tensor.
In 1-dimension strain is
 = ∆L/L (17)
where ∆L is the amount of length change and L is the length. The strain  is dimensionless.
An elastic solid has stress σ proportional to strain

σ = E (18)

and the coefficient E is Young’s modulus. However, a more realistic model takes into
account additional degrees of freedom (compression vs shear stress and strain). There are
additional moduli such as the bulk modulus. Stress and the associated moduli are in MKS
units of Pa (pascals) which is equivalent to N/m2 or J/m3 .

5.2 Deviations from linear elasticity

When pulled too far, the material can flow or be ductile or plastic instead of elastic. This is
why large solid asteroids or planets tend to be round. The material can also break. Tensile
yield strengths are often not the same as compressive yield strengths.
A material can heat up when stretched or compressed. With a velocity dependent or
strain rate dependent force, the material is described as viscoelastic.

6 Summary
• Hooke’s law F = −kx with positive spring constant k.
p
• Harmonic motion. Angular frequency ω = k/m. Period P = 2π/ω.

• How to find the solution for harmonic motion at later times from initial conditions.

19
 p
• The pendulum’s angular frequency ω = g/L for small amplitude motions.

• Driven and damped harmonic motion.

• Springs in parallel and in series.

• Elastic force in a solid and how this is related to interatomic forces.

• Practicing making force diagrams!

7 Props
Silly putty and a marble. Two slinkies.
Force chain movie called fromsolidtol.gif

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