Physical Chemistry Third Edition

14.2 Classical Mechanics 625

division byeλtgives thecharacteristic equation

λ^2 −

k

m

(14.2-20)

There are two solutions to this equation:

λ 1 i

√

k

m

(14.2-21a)

λ 2 −i

√

k

m

(14.2-21b)

whereiis theimaginary unit, defined to equal

√

−1. Each of these values ofλgives a

solution to the equation, so we have two solutions to the differential equation. Alinear

combination(sum of functions with constant coefficients) of the two solutions is also

a solution to the equation.

Exercise 14.1

The linear combination

c 1 eλ^1 t+c 2 eλ^2 tc 1 ei

√k/mt

+c 2 e−i

√k/mt

(14.2-22)

wherec 1 andc 2 are constants is a general solution to Eq. (14.2-14), since it contains two arbitrary

constants. Use theidentity(an equation that holds for any value ofα)

eiαcos(α)+isin(α) (14.2-23)

to show that this solution is identical to the solution in Eq. (14.2-15). Obtain the relation between

c 1 ,c 2 ,A, andB.

To make the general solution in Eq. (14.2-15) apply to a particular case we must

assign values toAandB. We useinitial conditionsto do this. One choice for the initial

conditions is to specify values of the position and velocity at the initial time. Let us

consider the case that

x(0)x 0 (14.2-24a)

vx(0) 0 (14.2-24b)

wherex 0 is a specified constant. Since cos(0)1 and sin(0)0, the first condition

requires thatBx 0 and the second condition requires thatA0. The specific solution

that applies to our initial conditions is now

x(t)x 0 cos(

√

k/mt) (14.2-25)

vx(t)−

√

k/m x 0 sin(

√

k/mt) (14.2-26)

The constantx 0 is the largest magnitude thatxattains and is called theamplitudeof

the oscillation. You can see thatxandvxare now determined for all values of the time,

both positive and negative. This is a characteristic of classical equations of motion. We

say that classical mechanics isdeterministic, which means that the classical equations

of motion determine the position and velocity of any particle for all time if the initial

conditions are precisely specified.