1
Linear Harmonic Oscillator
Consider the case of small energies, when the second derivative of
U
(
x
) at
x
= 0 is a
constant
k
. What is the nature of the motion? From the last class, the potential energy is
U
(
x
) =
kx
2
/
2, where
k
=
d
2
U
dx
2
0
. Since the force is the negative gradient of
U
,
F
(
x
) =

kx
(a
linear restoring force
), and Newton’s second law becomes:
m
¨
x
=

kx
(1)
or
¨
x
+
ω
2
0
x
= 0
(2)
where we have introduced
ω
0
=
k/m
. How do we solve this equation?
The general form of this equation is a
secondorder ordinary di
ff
erential equation
(
y
+
ay
+
by
=
f
(
x
)) that you will learn how to solve in a ODE class. Luckily, since
y
= 0 and
f
(
x
) = 0, we can take advantage of a special solution. First, multiply both sides by 2 ˙
x
:
2 ˙
x
¨
x
=

2
ω
2
0
x
˙
x
(3)
and integrate
˙
x
2
=

ω
2
0
x
2
+
C
(4)
where
C
=
ω
2
A
2
from initial conditions. This leads to
dx
√
A
2

x
2
=
ω
dt
(5)
which can be integrated to get
arcsin
x
A
=
ω
0
t
+
φ
(6)
thus
x
=
A
sin(
ω
0
t
+
φ
)
(7)
The two arbitrary constants must be determined from initial conditions.
ω
0
is the angular
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 Spring '10
 RogerMelko
 Energy, Kinetic Energy, Potential Energy, SMALL OSCILLATIONS, d2 U dx2

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