Environmental Engineering Reference
In-Depth Information
The Second Law specifies the equation of motion:
m
d
2
x
d
t
2
=−
kx
(3.21)
for which the general solution is
=
+
x(t)
A
sin
(ωt)
B
cos
(ωt),
(3.22)
m
is the angular frequency of
the oscillation. If you cannot already see it, then you should check that Eq. (3.22)
is a solution to Eq. (3.21) and that you agree with the expression for
ω
. Thus a
simple harmonic oscillator is characterised by an oscillation of constant frequency.
We are at liberty to define
x(
0
)
where
A
and
B
are arbitrary constants and
ω
=
=
0, that is we put the particle at the origin at time
zero. In which case, we have
B
=
0and
x(t)
=
A
sin
(ωt).
(3.23)
We can deduce an expression for the potential energy of a harmonic oscillator
by computing the work done in going from
x
=
x
A
to
x
=
x
B
x
B
2
k
x
A
−
x
B
.
1
W
AB
=−
kx
d
x
=
(3.24)
x
A
As usual we are free to select the point where the potential energy is zero, and we
choose the origin
x
=
0. So, setting
x
A
=
=
0 and defining
U(
0
)
0 we obtain
1
2
kx
B
.
W
AB
=
U(x
A
)
−
U(x
B
)
=−
U(x
B
)
=−
(3.25)
Thus, the potential energy of a harmonic oscillator is given by
1
2
kx
2
.
U(x)
=
(3.26)
The harmonic oscillator constitutes a conservative system and we can write the
mechanical energy:
1
2
mv
2
1
2
kx
2
,
=
+
E
(3.27)
where
v
is the speed of the particle. We can verify that this is a conserved quantity
by direct substitution of the solution (Eq. (3.23)) into the energy equation, i.e.
1
2
mω
2
A
2
cos
2
ωt
1
2
kA
2
sin
2
ωt
1
2
kA
2
.
E
=
K
+
U
=
+
=
(3.28)
Example 3.2.3
Determine the force acting on a particle that moves with a potential
energy U(x)
1
2
kx
2
αx
4
.
=
+
