Physical Chemistry , 1st ed.

which can be rearranged algebraically to yield the second-order differential

equation

d

dt

(^2) x
 2 m
kx 0
This differential equation has the general solution x(t) Asin tBcos
t,where Aand Bare constants characteristic of the particular system (de-
termined, for example, by the initial position and velocity of the oscilla-
tor) and


m
k

1/2
b.For Lagrange’s equations of motion, we need the kinetic energy Kand the
potential energy V. Both are the classical expressions
K^1
2
md
d
x
t

2
^1
2
mx ̇^2

V^1

2

kx^2

The Lagrangian function Lis thus

L

1

2

mx ̇^2 

1

2

kx^2

The Lagrange equation of motion for this one-dimensional system is



d

d

t





L

x ̇





L

x

 0

where equation 9.7 has been rewritten to equal zero. Recalling that ̇xis the

derivative ofxwith respect to time, we can take the derivative ofLwith re-

spect to ̇xas well as the derivative ofLwith respect to x. We find that







L

x ̇

mx ̇ 





L

x

kx

Substituting these expressions into the Lagrange equation of motion, we get



d

d

t

(m ̇x) kx 0

Since mass does not change with time, the derivative with respect to time af-

fects only ̇x. This expression then becomes

m

d

d

t

( ̇x) kx 0

which can be rearranged as

d

dt

(^2) x
 2 m
kx 0
This is the exact same second-order differential equation found using
Newton’s equations of motion. It therefore has the same solutions.
c.For Hamilton’s equation of motion, in this example the general coordinate
qis simply x, and ̇qequals ̇x. We need to find the momentum as defined by
equation 9.10. It is
pmx ̇
246 CHAPTER 9 Pre-Quantum Mechanics