Physical Chemistry Third Edition

626 14 Classical Mechanics and the Old Quantum Theory

Exercise 14.2

Show that the functions in Eq. (14.2-25) and Eq. (14.2-26) correspond to the given initial

conditions.

Figure 14.2a shows the position of the harmonic oscillator as a function of time, and

Figure 14.2b shows the velocity as a function of time. This motion is calleduniform

harmonic motion.Itisaperiodic motion, repeating the same pattern over and over

again as time passes. The length of time required for the oscillator to go from a given

state to the next repetition of that state is called theperiodof the oscillation and is

denoted byτ. It is the length of time required for the argument of the cosine function

in Eq. (14.2-25) and the sine function in Eq. (14.2-26) to change by 2π:

√

k

m

τ 2 π (14.2-27)

or

τ 2 π

√

m

k

(14.2-28)

Thefrequencyνis the number of oscillations per second and equals the reciprocal of

the period:

ν

1

τ



1

2 π

√

k

m

(14.2-29)

The frequency is larger if the force constant is larger, and smaller if the mass is larger.

The unit of frequency is the reciprocal second, called the hertz (Hz).

(b)

(a)

t

0

x

0

vx

t

Figure 14.2 The Behavior of a Har-

monic Oscillator.(a) The position as

a function of time according to classical

mechanics. (b) The velocity as a function

of time according to classical mechanics.

EXAMPLE14.1

A single covalent bond is similar to a spring with a force constant near 500 N m−^1. Estimate

the frequency of oscillation of a hydrogen atom at one end of such a spring with the other

end held fixed.

Solution

ν

1

2 π

√

500Nm−^1

1. 674 × 10 −^27 kg

 8. 7 × 1013 Hz

This frequency is similar to that of a molecular vibration.

The kinetic energyK of our harmonic oscillator is

K 

1

2

mv^2 x (14.2-30)

From Eq. (14.2-26) the kinetic energy is given as a function of time by

K 

1

2

mv^2 x

1

2

m

k

m

x^20

[

sin

(√

k

m

t

)] 2



k

2

x^20 sin^2

(√

k

m

t

)

(14.2-31)