chem 373- harmonic oscillator...classical

8/3/2019 Chem 373- Harmonic oscillator...classical

1/14

Harmonic oscillator...classical

Let us consider a particle of mass m attached to a spring

Equilibrium

x=0,t=0

o

x

Stretchx=xo

compressx=-xoxo

xo

At the beginning at t = o the particle is at equilibrium,

that is no particle is working at it , F = 0,


2/14


In general F = -k x . The force propotional to

displacement and pointing in opposite directiono

k is the force constant of the spring

Equilibrium

x=0,F=0

o

x

xo

xo

xo F=-kxo

xo

F= kxo


3/14

We might consider as an other example two particles

attached to each side of a spring

re

A B

F= 0 Equilibriumr = re

Case I: Equilibrium

A B

F= -kx Stretchr = re+x

Case II: Stretch

r = re+x



4/14


re-x

A B

F= -k(-x) Equilibrium

r = re

Case III: Compress

x

Again we have that the force F is proportionalto the displacement x

and pointing in the opposite direction

F = - k x


5/14


6/14


Let us look at this solution

(a) for or t = o we have x

for or t =2

x

o

o

k

mt

b km

t mk

A

= =

=

=

0 0

2

( )

( )

( )

( )

c km

t mk

o

dk

m tm

k A

e

k

m t

m

k o

for or t = x

for3

or t =3

x

for or t = 2 x

o

o

o

=

=

=

=

=

=

2 2

2


7/14

Equilibrium

x=0,t=0

o

x

Stretch

x=xo

compress

x=-xoxo

xo

position x = A sin (k

mt )


A

x

t

2

m

k

m

k

3

2

m

k

2 mk

-A


8/14

Ak

m

v

t

2

mk

m

k

3

2

m

k

2 mk

-Am

k

velocity v =dx

dt= A

k

mcos (

k

mt )



9/14

Ak

m

v

t

2

m

k

m

k

3

2

m

k

2m

k

-Am

k

Force = - k x = -Ak

msin (

k

mt )


Equilibrium

x=0,F=0

o

x

xo

xo

xo F=-kxo

xo

F= kxo


10/14


It

m

k

as

k

m

follows that the time to complete one cycle is

t

a consequence one can complete

=t

cycles per time unit

cycle

cycle

=

=

2

1 1

2

The frequency is often written

as =2

is referred to as the circular frequency.

We clearly have

=

k

m

where


11/14


We might also look at the kinetic energy

T =1

2t

t

t

mv m Ak

m

k

m

T mA km

km

T A kk

m

22

2 2

2 2

1

2

12

1

2

=

=

=

cos

cos

cos

t

2

m

k

m

k

3

2

m

k

2m

k1

2A2k

T


12/14

Harmonic oscillator...classicalWhat about potential energy

V(x) ??

We always have

F = -dV( x)

d x

F = -k x = -dV( x)

d x

dV

d xd x = V( x) - V(o)

= x d x = 12

x

V( x) = 12

x

x

x 2

2

Thus

or

k k o

Thus

k

0

0

V(x) = 1/2k2x2V

x-A2 A2k1 > k2

E

V(x) = 1/2k1x2

-A1 A1

t

2

m

k

mk

3

2

m

k

2m

k1

2 A2

k

V

V =1

2A2k sin2

k

mt


13/14


E T V A k km

A k km

E A k

We

= + =

+

=

12

12

1

2

2 2 2 2

2

cos cost t

note total energy independent of t

V(

x) = 1/2k

2x2

V

x-A2 A2k1 > k2

E

V(x) = 1/2k1

x2

-A1 A1


14/14

Harmonic oscillator...classicalFrom

k

m m

Thus

Note

depends

the relation

=

1

2 k = 4

E = 2 m A

that the amplitude A

A =2E

k

on E and k. For agiven E the smaller k the

larger A.

Note that the frequency isindependent of A

2

2 2

V(x) = 1/2k2x2V

x-A2 A2k1 > k2

E

V(x) = 1/2k1x2

-A1 A1

chem 373- harmonic oscillator...classical

Documents