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Damping and Driving Forces
Lachlan
August 2024

1
1
...
The gravitational force that
accelerates the object to the ground, given by mg, and a force of drag that slows the object
down, given by −bv
...

Now lets add together the forces and use Newton’s second law
...

mg − bv = m

dv
dt

F = ma is written in a slightly different form
...
Keep in mind that other
than v and t, everything else are constants
...
The left
hand side is
Z
Z
dv
m
dv
=
(3)
mg
b
b
v
g− m
b −v
m
mg
= − ln |
− v| + C
(4)
b
b
Which is just a simple u-sub
...
Note that the two constants of integration combine into
one big C
...
Lets determine A
...

Plugging 0 into v(t), we have
v(0) =

mg
mg
− A = 0 =⇒ A =
b
b

(7)

That means the velocity can be modelled by the function
v(t) =

b
mg
(1 − e− m t )
b

(8)

To know the terminal speed (the speed at which the object will stop accelerating in either
direction), we simply need to evaluate the limit as t → ∞, or
lim

t→∞

b
mg
mg
(1 − e− m t ) =
b
b

(9)

Since the exponential vanishes at infinity, leaving behind the linear term
...


1
...
Instead, quadratic air resistance is needed
...


2
2
...
Using
Newton’s Laws, we arrive at
m¨
x + bx˙ + kx = 0

(10)

This time, it is advantaguous to solve for the displacement x from the equillibrium point directly
rather than solve for the velocity first
...
We now take its derivative and obtain
x˙ = Aαeαt

(12)

2 αt

(13)

x
¨ = Aα e

We use these results to substitute into (10), leaving us with the simple task of finding the
coefficients A and α
Aeαt (mα2 + bα + k) = 0
2

(14)

We arrive at the quadratic equation
mα2 + bα + k = 0

(15)

Which has the solution
α=−
Where we defined the parameter ω0 ≡

q

p
b
± γ 2 − ω2
2m

k
m

Ω≡

and γ ≡
q

b
2m
...
If γ 2 > ω02 , then the equation of motion will be
x = e−ηt (Aeω0 t + Be−ω0 t )

(20)

Where A and B are constants determined by the initial conditions
...
If γ < η , Ω will be imaginary
...
This is known as underdamping, since the particle oscillates for a while before settling down
...
2

Driving Forces

Lets consider the previous system, but this time there exists an external driving force on the
particle
...
Using Newton’s
Laws, we arrive at the solution that
m¨
x + b¨
x + kx = F0 cos ω1 t
q
k
b
,γ≡m
and F1 ≡ Fm0
...
Assuming underdamping gives us
x0 = Ae−ηt cos(Ω′ t − ϕ)

(24)

Now we seek a particular solution
...
Immediately, we notice that
ζ = ω1
B=

(27)

F
2(ω02 − ω12 + iγω1 )

(28)

Now we find that our particular solution xp becomes
xp =
If we define the quantity R ≡

p

2(ω02

F
(eiω1 t − e−iω1 t )
− ω12 + iγω1 )

(29)

ω02 − ω12 + (2γω0 )2 and simplify (29), we arrive at

xp =

F ω02 − ω12
2ω1
(
cos ω1 t +
sin ω1 t)
R
R
R

(30)

A little rearranging and particular choice of coefficients ϕ2 and ν gives
xp =

F
cos(νt − ϕ2 )
R

(31)

Putting everything together we finally have
x = Ae−ηt cos(Ω′ t − ϕ) +

F
cos(νt − ϕ2 )
R

(32)

Notice how the amplitude of oscillation is proportional to R−1 , and R gets smaller when the
difference between the natural frequency of the object, ω0 and the driving frequency ω1 gets
smaller
...
This is phenomenon known as resonance
...
We call the first term the ”transient” and the second term the ”attractor”, since
the system gets ”attracted” towards the second term as time goes on
...
3

Two Coupled Oscillators

Imagine you have four identical masses m attached to each other by identical springs with spring
constant k, where the outermost masses are pinned down and cannot move, while the middle
two masses are confined to move along the x-axis
...
The displacement of the spring between any two masses can be
found by the difference of their positions
...

m¨
x1 = k(x1 − x0 ) − k(x2 − x1 )

(33)

m¨
x1 = k(x2 − x1 ) − k(x3 − x2 )

(34)

The first terms are positive because of Newton’s third law
...
We know that x0 = x3 = 0, so
using this fact and simplifying (33) and (34), we arrive at
x
¨1 = −ω 2 (2x1 − x2 )
2

x
¨2 = −ω (2x2 − x1 )
4

(35)
(36)

Where ω ≡

q

k
m
...
Lets guess a solution in the form
xn = An eηn t =⇒ x
¨n = An η 2 eiηt

(37)

Substituting this into (35) and (36), we have
−A1 η 2 eiηt = −ω 2 (2A1 eiηt − A2 eiηt )
2 iηt

−A2 η e

2

= −ω (2A2 e

iηt

− A1 e

iηt

)

(38)
(39)

We seek to find the coefficients An and η
...
Rewriting in vector
form gives

 
 
−2ω 2
ω2
A1
2 A1
=
−η
(42)
ω2
−2ω 2 A2
A2
To find η, we simply have to take the determinant of the matrix and set it to 0



−2ω 2 + η 2
ω2


2
2
2 = 0

ω
−2ω + η
η 2 = ±ω 2 + 2ω 2
√
η = {±ω, ± 3ω}

(43)
(44)
(45)

To find A1 and A2 is trivial, for η 2 = ω 2 , A1 = A2 = 1
...
Using
the identity that A(eiωθ + e−iωθ ) = A2 cos(ωθ + ϕ), the equations of motion are
√

x1 = B1 cos(ωt + ϕ1 ) + B2 cos 3ωt + ϕ2
(46)
√

x2 = B1 cos(ωt + ϕ1 ) − B2 cos 3ωt + ϕ2
(47)
Where Bn = A2n
...


2
...
Generally, the equation of motion on any individual mass will
be
x
¨n = −ω 2 (2xn − xn+1 − xn−1 )

(48)

If we make the same assumptions as (37) and follow the same steps, we arrive at the eigenequation
 2




−ω
2ω 2 −ω 2
0

...

0 

  A1




...


...

2

...


...

=
−η
(49)





...


...


...




...
−ω
2ω
−ω
0
AN
AN
0

...
If we know N , we can then solve for the eigenvectors and
eigenvalues to obtain the equations of motion and thus, the normal modes

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