5.61 Fall 2007 Lectures #12-15 page 1
THE HARMONIC OSCILLATOR
• Nearly any system near equilibrium can be approximated as a H.O.
• One of a handful of problems that can be solved exactly in quantum
mechanics
examples
m1 m2 B (magnetic
field)
A diatomic molecule µ
(spin
magnetic
moment)
E (electric
field)
Classical H.O.
m
k
X0 X
Hooke’s Law: (
f = −k X − X 0 ≡ −kx)
(restoring force)
d2x d2x ⎛ k ⎞
f = ma = m 2 = −kx ⇒ + x=0
dt dt 2 ⎜⎝ m⎠⎟
,5.61 Fall 2007 Lectures #12-15 page 2
Solve diff. eq.: General solutions are sin and cos functions
k
() ( )
x t = Asin ω t + B cos ω t ( ) ω=
m
or can also write as
() (
x t = C sin ω t + φ )
where A and B or C and φ are determined by the initial conditions.
e.g. ()
x 0 = x0 ()
v 0 =0
spring is stretched to position x 0 and released at time t = 0.
Then
() ()
x 0 = A sin 0 + B cos 0 = x0 () ⇒ B = x0
dx
()
v 0 =
dt
()
= ω cos 0 − ω sin 0 = 0() ⇒ A=0
x=0
So ()
x t = x0 cos ω t ( )
k
Mass and spring oscillate with frequency: ω =
m
and maximum displacement x0 from equilibrium when cos(ωt)= ±1
Energy of H.O.
Kinetic energy ≡ K
2
1 1 ⎛ dx ⎞ 1 1
( ) ( )
2
K = mv 2 = m ⎜ ⎟ = m ⎡⎣ −ω x0 sin ω t ⎤⎦ = kx02 sin 2 ω t
2 2 ⎝ dt ⎠ 2 2
Potential energy ≡ U
dU 1 1 2
()
f x =−
dx
⇒ U = − ∫ f x dx =() ∫ ( kx )dx =
2
kx 2
=
2
( )
kx0 cos 2 ω t
, 5.61 Fall 2007 Lectures #12-15 page 3
Total energy = K + U = E
1 2 1 2
E=
2
( ) ( )
kx0 ⎡⎣sin 2 ω t + cos 2 ω t ⎤⎦ E= kx
2 0
x (t )
x 0(t )
0 t
-x0(t)
U K
1 2
kx E
2 0
0 t
Most real systems near equilibrium can be approximated as H.O.
e.g. Diatomic molecular bond A B
X
U
X
X0 A + B separated atoms
equilibrium bond length
THE HARMONIC OSCILLATOR
• Nearly any system near equilibrium can be approximated as a H.O.
• One of a handful of problems that can be solved exactly in quantum
mechanics
examples
m1 m2 B (magnetic
field)
A diatomic molecule µ
(spin
magnetic
moment)
E (electric
field)
Classical H.O.
m
k
X0 X
Hooke’s Law: (
f = −k X − X 0 ≡ −kx)
(restoring force)
d2x d2x ⎛ k ⎞
f = ma = m 2 = −kx ⇒ + x=0
dt dt 2 ⎜⎝ m⎠⎟
,5.61 Fall 2007 Lectures #12-15 page 2
Solve diff. eq.: General solutions are sin and cos functions
k
() ( )
x t = Asin ω t + B cos ω t ( ) ω=
m
or can also write as
() (
x t = C sin ω t + φ )
where A and B or C and φ are determined by the initial conditions.
e.g. ()
x 0 = x0 ()
v 0 =0
spring is stretched to position x 0 and released at time t = 0.
Then
() ()
x 0 = A sin 0 + B cos 0 = x0 () ⇒ B = x0
dx
()
v 0 =
dt
()
= ω cos 0 − ω sin 0 = 0() ⇒ A=0
x=0
So ()
x t = x0 cos ω t ( )
k
Mass and spring oscillate with frequency: ω =
m
and maximum displacement x0 from equilibrium when cos(ωt)= ±1
Energy of H.O.
Kinetic energy ≡ K
2
1 1 ⎛ dx ⎞ 1 1
( ) ( )
2
K = mv 2 = m ⎜ ⎟ = m ⎡⎣ −ω x0 sin ω t ⎤⎦ = kx02 sin 2 ω t
2 2 ⎝ dt ⎠ 2 2
Potential energy ≡ U
dU 1 1 2
()
f x =−
dx
⇒ U = − ∫ f x dx =() ∫ ( kx )dx =
2
kx 2
=
2
( )
kx0 cos 2 ω t
, 5.61 Fall 2007 Lectures #12-15 page 3
Total energy = K + U = E
1 2 1 2
E=
2
( ) ( )
kx0 ⎡⎣sin 2 ω t + cos 2 ω t ⎤⎦ E= kx
2 0
x (t )
x 0(t )
0 t
-x0(t)
U K
1 2
kx E
2 0
0 t
Most real systems near equilibrium can be approximated as H.O.
e.g. Diatomic molecular bond A B
X
U
X
X0 A + B separated atoms
equilibrium bond length
