Surface tension
Weight (mass x gravity)
Impedance
Flow
Water Bubbles
Dripping Tap
Adhesion (surface tension)
Buoyancy
Flow
Impedance
Figure 4.4
Quantisation of flow.
SECTION 4.2
Simple Harmonic Oscillators
We have already mentioned simple harmonic motion while talking about how
points of matter behave in a vibrating object, more properly described as the
44
Oscillations
free vibrations of a lumped (discrete) system with a limited number of degrees
of freedom and without exciting force. The main condition of simple harmonic
motion occurs when the force acting on a body, and thus accelerating it, is
proportional to and in the opposite direction from its displacement from the
rest point. This is determined by a differential equation
A
d
2
x
dt
2
=
−
Bx
(4.2)
where
B
is a constant of proportional restoration,
x
is the displacement,
A
is a
variable that determines the period of movement (usually mass), and
t
is time.
Solving the equation gives us a formula for the frequency of the system, and we
shall briefly consider two common examples now, the undamped mass-spring
and the pendulum.
Frequency of a Mass on Spring
A familiar school physics experiment is to calculate the frequency of a mass on
a spring. In an idealised form it has only one degree of freedom: it can go up
and down, but not side to side. Consider a spring and mass, here on Earth,
at some time when it is just sitting there at rest. Gravity acts on the mass
giving it weight, a force acting downwards. This force
F
is the mass
m
times
the gravitational constant
g
= 9
.
8, so
F
=
m
×
g
, and if the mass is 10kg then
it creates a force (in Newtons N) of 10
×
9
.
8 = 98N. Another force is acting in
the opposite direction, keeping the system in equilibrium. This is the support
offered by the spring (and whatever the spring is attached to) in its rest posi-
tion. Now, the spring has an elastic property, so if it is stretched or compressed
some distance
x
by a force it will produce an opposing, restoring force pro-
portional to that distance. The springiness is a constant
k
, measured in N
/
m.
Hooke’s law says that in a linear system, the restoring force is proportional to
the displacement of the mass, and acts in a direction to restore equilibrium, so
we write
F
=
−
kx
. If the spring was initially unloaded and adding a mass
m
kg
caused it to stretch by
x
m we know its springiness
k
to be
−
mg/x
since the
force created by gravity
mg
must equal the force from the spring
−
kx
holding
up the mass. Let’s say our 10kg mass caused the spring to move 1cm, which
makes
k
= 98N
/
0
.
01m = 9800N
/
m. After working this out we can ignore gravity,
because, maybe surprisingly, the frequency of a spring and mass doesn’t depend
on it. If you take the spring and mass out into space it will still oscillate at
the same frequency. But what will that frequency be? Well, an oscillating mass
must move, and a moving mass requires a force to either make it speed up or
slow down. From Newton’s second law,
F
=
m
×
a
, where
a
is the acceleration
in ms
−
2
and
m
is the mass. Assuming for a moment there are no losses, in
keeping with conservation laws the forces must balance at all times. So at any
time we have an equation
ma
=
−
kx
. Remember that acceleration is the rate
of change of velocity, and velocity is the rate of change of position. This leads
us to a differential equation
m
d
2
x
dt
2
=
−
kx
(4.3)
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