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1 Differential Equations and Newton's Laws of Motion

1 Differential Equations and Newton's Laws of Motion

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140



Mathematics for Physical Chemistry



The acceleration a is the rate of change of the velocity:

d2 r

dv

d2 x

d2 y

d2 z

= 2 = i 2 +j 2 +k 2 .

dt

dt

dt

dt

dt

(12.3)

Equations of motion are obtained from Newton’s three laws

of motion: 1



a = iax + ja y + kaz =



1. A body on which no forces act does not accelerate.

2. A body acted on by a force F accelerates according to

F = ma,



(12.4)



where m is the mass of the object and a is its acceleration.

3. Two bodies exert forces of equal magnitude and opposite

direction on each other.

Classical mechanics (sometimes called Newtonian

mechanics) is primarily the study of the consequences of

Newton’s laws. We now accept them as approximations to

reality that are accurate for large objects moving at speeds

much less than the speed of light. The first law is just a

special case of the second, and the third law is primarily

used to obtain forces for the second law, so Newton’s second

law is the most important equation of classical mechanics.

We can write an equation of motion if the acceleration

of a particle is known as a function of time. If the particle

moves only in the z direction:

az = az (t).



(12.5)



We equate the time derivative of the velocity to this known

function and obtain an equation of motion:

dvz

= az (t).

dt



(12.6)



To solve Eq. (12.6), we multiply both sides by dt and

perform a definite integration from t = 0 to t = t1 .

vz (t1 ) − vz (0) =



t1

0



dvz

dt



dt =



t1

0



az (t)dt.



(12.7)



The result of this integration gives vz as a function of time,

so that the position obeys a second differential equation

dz

= vz (t).

dt



(12.8)



1 Isaac Newton (1642–1827) was a great English physicist who deduced



the law of gravity and three laws of motion from mathematical analysis

of observations on the motions of planets around the sun.



A second integration gives the position as a function of

time:

t2



z z (t2 ) − z(0) =



0

t2



=



0

t2



=



0



dz

dt1



t2



dt1 =



vz (0) +

vz (0)dt1 +



0



t1

0



az (t)dt dt1



t2

0



vz (t1 )dt1



t1

0



az (t)dtdt1 .

(12.9)



There are inertial navigation systems used on submarines

and space vehicles that measure the acceleration as a

function of time and perform two numerical integrations

in order to determine the position of the vehicle.

Example 12.1. At time t = 0, a certain particle has z(0) = 0

and vz (0) = 0. Its acceleration is given as a function of time

by

az (t) = a0 e−t/b ,

where a0 and b are constants.

(a) Find vz as a function of time.

The velocity is the antiderivative of the given

acceleration function plus a constant v0 :

vz (t) = −a0 be−t/b + v0

= −a0 be−t/b + a0 b = a0 b(1 − e−t/b ),

since it was specified that vz (0) = 0, the constant

v0 = a0 b.

(b) Find z as a function of time.

The position is the antiderivative of the velocity plus

a constant z 0 :

z(t) = a0 b2 e−t/b + a0 bt + z 0

= a0 b2 e−t/b + a0 bt − a0 b2 ,

since it was specified that z(0) = 0, the constant z 0 is

equal to a0 b2 .

(c) Find the speed and the position of the particle at

t = 30.0 s if a0 = 10.0 m s−2 and if b = 20.0 s. At

t = 30.0 s

vz (30.0 s) = (10.0 m s−2 )(20.0 s)(1 − e−1.50 )

= 155 m s−1 ,

z(30.0 s) = (10.0 m s−2 )(20.0 s)2 e−1.50

+(200.0 m s−1 )(30.0 s)

−(10.0 m s−2 )(20.0 s)2 = 2890 m.

(d) Find the limiting value of the speed as t → ∞:

lim vz (t) = a0 b = 200. ms−1



t→∞



CHAPTER | 12 Differential Equations



141



Exercise 12.1. An object falling in a vacuum near the

surface of the earth experiences a gravitational force in the

z direction given by

Fz = −mg,

where g is called the acceleration due to gravity and is equal

to 9.80 m s−2 .2 This corresponds to a constant acceleration

az = −g.

Find the expression for the position of the particle as a

function of time. Find the position of the particle at time

t = 1.00 s if its initial position is z = 10.00 m and its initial

velocity is vz = 0.

Equations of motion can also be obtained if the force on

a particle is known as a function of position. We consider

some of these in the next sections.



12.2 HOMOGENEOUS LINEAR

DIFFERENTIAL EQUATIONS WITH

CONSTANT COEFFICIENTS

An ordinary linear differential equation with constant

coefficients has the properties:

1. Ordinary means that it contains no partial derivatives.

2. Linear means that it contains the function and its

derivatives only to the first power.

3. Constant coefficients mean that the quantities multiplying the dependent variable and its derivatives are

constants.

We first consider homogeneous equations, which means

that there is no term that does not contain the function or

its derivatives. The equation can be written

a0 y + a1



dy

d2 y

d3 y

+ a2 2 + a3 3 + · · · = 0.

dx

dx

dx



(12.10)



12.2.1 The Harmonic Oscillator

Consider an object that is suspended at the end of a

coil spring whose other end is stationary. The object can

oscillate in the z direction. Let z = 0 when the spring has its

equilibrium length. If we can ignore gravity, the only force

on the object is due to the spring and is given to a good

approximation by Hooke’s law,3

Fz = −kz,



(12.11)



2 The acceleration due to gravity actually depends slightly on latitude,



having a larger value near the poles. This value applies at the latitude of

Washington, DC, and Seoul, South Korea.

3 After Robert Hooke, 1635–1703, one of Newton’s contemporaries and

rivals.



where k is a constant called the spring constant. The

negative sign indicates a negative force when z is positive

and vice versa, so that the force pushes the mass toward its

equilibrium position. A real object on a spring is difficult

to analyze exactly, because the spring also moves as it

stretches and compresses and because Hooke’s law is an

approximation that fails when the spring is stretched beyond

its elastic limit and when it is compressed enough that the

coils interfere with each other. We now define the harmonic

oscillator, which is a model system that approximately

represents the object on the spring. A model system is

a hypothetical system (existing only in our minds) that

has some properties in common with a real system, but

is enough simpler to allow mathematical analysis. The

harmonic oscillator is defined by assuming that the spring

has no mass and that Hooke’s law is exactly obeyed, even

if z has a large magnitude.



The Equation of Motion of the Harmonic

Oscillator

Newton’s second law gives the equation of motion for the

harmonic oscillator:

d2 z

= −kz.

(12.12)

dt 2

We divide by m and move all terms to the left-hand side of

the equation to obtain

m



d2 z

k

(12.13)

+ z = 0.

dt 2

m

This is a homogeneous ordinary linear differential equation

with constant coefficients. We say that it is second order,

which means that the highest order derivative in the

equation is a second derivative.

There are two important facts about linear homogeneous

differential equations:

1. If z(t) satisfies the equation, then cz(t) is also a solution,

where c is a constant.

2. If z 1 (t) and z 2 (t) are two functions that satisfy the

equation, then the linear combination z 3 (t) is also a

solution:

z 3 (t) = c1 z 1 (t) + c2 z 2 (t),



(12.14)



where c1 and c2 are constants.



Solution of the Equation of Motion

A homogeneous ordinary linear differential equation with

constant coefficients can be solved as follows:

1. Assume the trial solution

z(t) = eλt ,

where λ is a constant.



(12.15)



142



Mathematics for Physical Chemistry



2. Substitute the trail solution into the equation and

produce an algebraic equation in λ called the

characteristic equation.

3. Find the values of λ that satisfy the characteristic

equation. For an equation of order n, there will be

n values of λ, where n is the order of the equation.

Call these values λ1 ,λ2 , . . . ,λn . These values produce n

versions of the trial solution that satisfy the equation.

4. Write the solution as a linear combination

z(t) = c1 eλ1 t + c2 eλ2 t + · · · + cn eλn t .



from knowledge of the state of the system at some initial

time.

We now apply the method to the solution of the equation

of motion of the harmonic oscillator. We substitute the trial

solution into Eq. (12.12):

k

d2 eλt

+ eλt = 0,

dt 2

m

k

λ2 eλt + eλt = 0.

m



(12.16)



Example 12.2. Find a solution to the differential equation



λ2 +



Division by eλx gives the characteristic equation.

λ2 + λ − 2 = 0.

The solutions to this equation are

λ = 1, λ = −2.



k

= 0.

m



y(x) = c1 e + c2 e



−2x



λ = ±i



k

m



1/2



+c2 exp −i



,



The solution in the previous example is a family of

solutions, one solution for each set of values for c1 and

c2 . A solution to a linear differential equation of order n

that contains n arbitrary constants is known to be a general

solution, which is a family of functions that includes almost

every solution to the differential equation. Our solution is a

general solution, since it contains two arbitrary constants. A

solution to a differential equation that contains no arbitrary

constants is called a particular solution.

Exercise 12.2. Find the general solution to the differential

equation

d2 y

dy

+ 2y = 0.

−3

dx 2

dx

We frequently have additional information that will

enable us to pick a particular solution out of a family

of solutions. Such information consists of knowledge

of boundary conditions and initial conditions. Boundary

conditions arise from physical requirements on the solution,

such as conditions that apply to the boundaries of the region

in space where the solution applies. Initial conditions arise



,



(12.20)





where i = −1, the imaginary unit.

The general solution is

z = z(t) = c1 exp +i



where c10 and c2 are constants.



(12.19)



The solution of the characteristic equation is



The solution to the differential equation is

x



(12.18)



Division by eλt gives the characteristic equation



d2 y

dy

− 2y = 0.

+

dx 2

dx

Substitution of the trial solution y = eλx gives the

equation

λ2 eλx + λ eλx − 2 eλx = 0.



(12.17)



k

m



1/2



k

m



t



1/2



t ,



(12.21)



where c1 and c2 are arbitrary constants. The solution

must be real, because imaginary and complex numbers

cannot represent physically measurable quantities. From

a trigonometric identity

eiωt = cos(ωt) + i sin(ωt),



(12.22)



we can write

z = c1 cos(ωt) + i sin(ωt) + c2 cos(ωt) − i sin(ωt) ,

(12.23)

where

k 1/2

ω=

.

(12.24)

m

We let c1 + c2 = b1 and i(c1 − c2 ) = b2 .

z = b1 cos(ωt) + b2 sin(ωt).



(12.25)



Exercise 12.3. Show that the function of Eq. (12.25)

satisfies Eq. (12.12).



CHAPTER | 12 Differential Equations



143



To obtain a particular solution, we require some Initial

conditions. We require one initial condition to evaluate each

arbitrary constant. Assume that we have the conditions at

t = 0:

z(0) = 0,

(12.26a)



by 2π. We denote the period (the length of time required

for one cycle of the motion) by τ :

2π = ωτ =

τ = 2π



vz (0) = v0 ,



(12.27)



where v0 is a constant. For our first initial conditions,

z(0) = 0 = b1 cos (0) + b2 sin (0) = b1 = 0, (12.28)

z(t) = b2 sin(ωt).



(12.29)



The velocity is given by

vz (t) =



dz

= b2 ω cos(ωt).

dt



(12.30)



From our second initial condition



k

m



m

k



1/2



1/2



τ,

.



(12.33)

(12.34)



The reciprocal of the period is called the frequency, denoted

by ν 4 :

ω

1

k

=

.

(12.35)

ν=

2π m



The frequency gives the number of oscillations per second.

The circular frequency ω gives the rate of change of the

argument of the sine or cosine function in radians per

second.

Example 12.3. A mass of 0.100 kg is suspended from

a spring with a spring constant k = 1.50 N m−1 . Find the

frequency and period of oscillation.



vz (0) = v(0) = b2 ω cos (0) = b2 ω.

This gives

b2 =



v0

.

ω



(12.31)



Our particular solution is

z(t) =



1

k

1

=

2π m



1

τ = = 1.62 s.

ν

ν =



v0

sin(ωt)

ω



(12.32)



as depicted in Figure 12.1.

The motion given by this solution is called uniform

harmonic motion. It is periodic, repeating itself over and

over. During one period, the argument of the sine changes



1.50 N m−1

= 0.616 s−1 ,

0.100 kg



The solution of the equation of motion of the harmonic

oscillator illustrates a general property of classical

equations of motion. If the equation of motion and the

initial conditions are known, the motion of the system is

determined as a function of time. We say that classical

equations of motion are deterministic.



The Vibration of a Diatomic Molecule

x



The vibration of a diatomic molecule can be modeled as a

harmonic oscillator. Since both nuclei move, the mass must

be replaced by the reduced mass,



v0

ω

1

τ =ν



2

2τ =ν



t



v0



0



μ=



τ







3τ t



FIGURE 12.1 The position and velocity of a harmonic oscillator as

functions of time.



m1m2

,

m1 + m2



(12.36)



where m 1 is the mass of one atom and m 2 is the mass of the

other atom. The frequency of vibration is

ν=



v



0



3

3τ =ν



1

ω

=







k

.

μ



(12.37)



Example 12.4. Calculate the frequency of vibration of a

hydrogen fluoride molecule. The force constant k is equal

to 966 N m−1 = 966 J m−2 .

4 In some fonts, the Greek letter ν (nu) and the italic letter v (vee) look



almost identical. Try not to be confused.



144



Mathematics for Physical Chemistry



We can use average molar masses, since fluorine occurs

with only one isotope and ordinary hydrogen is nearly

all one isotope. We first calculate the reduced mass times

Avogadro’s constant :

μNAv =



1.008 g mol−1 (18.998 g mol−1 )

1.008 g



mol−1



+ 18.998 g



mol−1



1 kg

1000 g



−4



= 9.572 × 10 kg mol−1

9.572 × 10−4 kg mol−1

μ=

= 1.589 × 10−27 kg

6.02214 × 1023 mol−1

1

ν =



=



1





966 kg m2 s−2 m−2

= 1.24 × 1014 s−1

1.589 × 10−27 kg



Exercise 12.4. The frequency of vibration of the H2

molecule is 1.3194 × 1014 s−1 . Find the value of the force

constant.



The Energy of a Harmonic Oscillator



1 2

1

kxmax =

1.50 N m−1

2

2

= 0.0108 N m = 0.0108 J.



1 2

mv .

2 z



(12.38)



According to classical mechanics, if the force on an object

depends only on its position, the force can be derived from

a potential energy function. For an object moving only in

the z direction

∂V

(12.39)

Fz = − ,

∂z

where V is the potential energy. From the equation for the

force, the potential energy of the harmonic oscillator must

be given by

1

V (z) = kz 2 .

(12.40)

2

The total energy is the sum of the kinetic energy and the

potential energy

2



sin2 (ωt) =



0.120 m



2



When z = 0, the energy is all kinetic energy and the velocity

has its maximum value:

E = Kmax =

vmax =



2E

=

m



1 2

mv ,

2 max

2 0.0108 J

= 0.465 m s−1 .

0.100 kg



Exercise 12.5. According to quantum mechanics, the

energy of a harmonic oscillator is quantized. That is, it can

take on any one of a certain set of values, given by



The kinetic energy of the harmonic oscillator is



E = K+V

1

1 v0

= mv02 cos2 (ωt) + k

2

2

ω



Example 12.5. A mass of 0.100 kg is suspended from

a spring with a spring constant k = 1.50 N m−1 . If its

amplitude of oscillation (the maximum value of z) is

0.120 m, find the energy of the oscillator. Find the maximum

velocity.

At the maximum value of z , the energy is all potential

energy:

E = Vmax =



966 J m−2

1.589 × 10−27 kg



K=



on the particles of a system can be obtained from a potential

energy function, the system will be conservative.



1 2

mv ,

2 0

(12.41)



where we have used the identity of Eq. (7) of Appendix

B. As a harmonic oscillator moves, the kinetic energy

rises and falls, but the potential energy changes so that the

total energy remains constant. If the energy of a system

remains constant during its motion, we say that the energy

is conserved, and that the system is conservative. There is

an important theorem of classical mechanics: If the forces



E = hν v +



1

,

2



where h is Planck’s constant, equal to 6.62608 × 10−34 J s,

ν is the frequency and v is a quantum number, which can

equal 0, 1, 2, … The frequency of oscillation of a hydrogen

molecule is 1.319 × 1014 s−1 . If a classical harmonic

oscillator having this frequency happens to have an energy

equal to the v=1 quantum energy, find this energy. What is

the maximum value that its kinetic energy can have in this

state? What is the maximum value that its potential energy

can have? What is the value of the kinetic energy when the

potential energy has its maximum value?



12.2.2 The Damped Harmonic

Oscillator—A Nonconservative

System

The damped harmonic oscillator is a harmonic oscillator

that is subject to an additional friction force that is

proportional to the velocity and in the opposite direction:

Ff = −ζ v = −ζ



dr

,

dt



(12.42)



where ζ is called the friction constant. An example of such

a force is the frictional force on a spherical object moving

relatively slowly through a viscous fluid. Since this force



145



cannot be derived from a potential energy, the system is not

conservative and its energy will change with time.

If the oscillator moves only in the z direction, the

equation of motion is

−ζ



dz

− kz = m

dt



d2 z

dt 2



.



z



CHAPTER | 12 Differential Equations



(12.43)



This equation is a linear homogeneous equation with

constant coefficients, so a trial solution of the form of Eq.

(12.15) will work. Substitution of the trial solution into the

equation gives



0



t



FIGURE 12.2 The position of a greater than critically damped harmonic

oscillator as a function of time.



−ζ λeλt − keλt = mλ2 eλt .

Division by meλt gives the characteristic equation

λ2 +



k

ζλ

+

= 0.

m

m



(12.44)



From the quadratic formula, the two solutions of this

equation are

ζ /m



ζ

λ1 = −

+

2m



2



− 4k/m



2



,



(12.45)



Figure 12.2 shows the position of a greater than critically

damped oscillator as a function of time for the case that

c1 = −c2 so that z = 0 at t = 0. Since the two terms decay

at different rates, the oscillator moves away from z = 0 and

then returns smoothly toward z = 0.

Example 12.6. From the fact that ζ , k, and m are all

positive, show that λ1 and λ2 are both negative in the case

of greater than critical damping, and from this fact, show

that

lim z(t) = 0,

t→∞



2



ζ /m − 4k/m

ζ



(12.46)

λ2 = −

2m

2

and the general solution to the differential equation is

z(t) = c1 eλ1 t + c2 eλ2 t .



(12.47)



ζ



λ2 = −

2m

Since ζ /m



2



Exercise 12.6.

equation.



does satisfy the differential



Greater than Critical Damping

There are three cases: In the first case, the quantity inside the

square root in Eq. (12.45) is positive, so that λ1 and λ2 are

both real. Since ζ , k, and m are all positive, this corresponds

to a relatively large value of the friction constant ζ ,

ζ /m



2



ζ

m

ζ



4k

,

m

4k

,

m



4km.



2



This case is called greater than critical damping. In

this case, the mass at the end of the spring does not

oscillate, but returns smoothly to its equilibrium position

of z = 0 if disturbed from this position. The values of the

constants c1 and c2 must be determined for a particular case.



− 4k/m



2



.



− 4k/m



0



the square root in the second term is real and the second

term is negative. Therefore λ2 must be negative,

ζ

+

λ1 = −

2m



ζ /m



2



− 4k/m



2



.



Since the second term is positive, we need to show that it is

smaller in magnitude than the first term. Square each term.

We must have

ζ2

4m 2



(12.48)



2



is larger than 4k/m

ζ /m



Show that eλ1 t



ζ /m



ζ /m



2



− 4k/m

.

4



Subtract ζ 2 /2m 2 from each side of this inequality. This does

not change the sense of the inequality

0







k

.

m



This is obviously correct, so λ1 must also be negative. Both

terms in (12.47) approach zero for large values of the time.



146



Mathematics for Physical Chemistry



Less than Critical Damping

The next case is that of small values of the friction constant

ζ , which is called less than critical damping. If

ζ

m



2



<



4k

,

m



ζ

+ iω,

2m

ζ

− iω,

λ2 = −

2m



λ1 = −



(12.49)

(12.50)



where we let

k



m



ζ

2m



2



ζ

2m



the quantity inside the square root in Eq. (12.45) is negative,

so that λ1 and λ2 are complex quantities:



ω=



is possible to construct an oscillating object such as a

galvanometer mirror or a two-pan balance beam that is very

nearly critically damped by a magnetic field. The condition

for critical damping is



2



.



(12.51)



The solution thus becomes

z(t) = c1 eiωt e−ζ t/2m + c2 e−iωt e−ζ t/2m .



(12.52)



Using Eq. (21) in Appendix B, we can write

z(t) = b1 cos(ωt) + b2 sin(ωt) e−ζ t/2m .



=



k

.

m



(12.54)



In the case of critical damping the two values of λ are equal

to each other,

ζ

,

(12.55)

λ1 = λ 2 = −

2m

so that

(12.56)

z(t) = c1 + c2 eλt = ceλt ,

where c is the sum of the two constants c1 and c2 and where

we drop the subscript on λ. A general solution for a secondorder linear equation contains two arbitrary constants, and

a sum of two constants does not constitute two separate

constants. Since we do not have a general solution, there

must be another family of solutions. We attempt additional

trial functions until we find one that works. The one that

works is

(12.57)

z(t) = teλt .



(12.53)



This shows z(t) to be an oscillatory function times an

exponentially decreasing function, giving the “ringing”

behavior shown in Figure 12.3.



Exercise 12.8. Substitute this trial solution into Eq.

(12.43), using the condition of Eq. (12.54), and show that

the equation is satisfied.

Our general solution is now

(12.58)



z



z(t) = c1 + c2 t eλt ,



where we omit the subscript on λ. The velocity is given by

vz (t) =

0



t



FIGURE 12.3 The position of a less than critically damped oscillator as

a function of time for the case that b2 = 0 and the initial conditions are

that z(0) = z 0 and that vz (0) = 0.



Exercise 12.7. If z(0) = z 0 and if vz (0) = 0, express the

constants b1 and b2 in terms of z 0 .



Critical Damping

The final case is that of critical damping. In this case

the quantity inside the square root in Eq. (12.45) is equal

to zero. This is not likely to happen by chance, but it



dz

= c1 λeλt + c2 eλt + c2 tλeλt .

dt



For any set of initial conditions, we can find the appropriate

values of c1 and c2 . The behavior of a critically damped

oscillator is much the same as that of Figure 12.2, with no

oscillation.

Example 12.7. Consider a critically damped oscillator

with λ = −1.00 s−1 . Assume that its initial position is

z(0) = 0.00 m and that its initial velocity is 1.00 m s−1 .

Find its position and velocity at t = 2.00 s.

In order for z 0 to equal 0.00 m , we must require that

c1 = 0.0 m. The velocity is then given by

vz (t) = c2 eλt + c2 tλeλt .

The velocity at t = 0 is

vz (0) = c2 = 1.00 m s−1 ,

z(t) = 1.00 m s−1



1.00 s−1 t



× exp −(1.00 s−1 )t .



CHAPTER | 12 Differential Equations



147



The position at t = 2.00 s is

z(2.00 s) = c2 te



We divide by m:



λt



F t

k

d2 z

.

(12.61)

+ z=

2

dt

m

m

The term F(t)/m is the inhomogeneous term.

A method for solving such an inhomogeneous

equation is:



= (1.00 m s−1 )(2.00 s)e−2.00 = 0.2707 m,

the velocity at t = 2.00 s is

vz (2.00 s) = 1.00 m s−1 e−2.00

+(1.00 m s−1 ) −1.00 s−1



2.00 s e−2.00



= −0.2707 m s−1 .

Figure 12.4 shows the position of the critically damped

oscillator in this example as a function of time.

0.4

0.35

0.3

0.25



Step 1. Solve the equation obtained by deleting the

inhomogeneous term. This homogeneous equation

is called the complementary equation, and the

general solution to this equation is called the

complementary function, z c .

Step 2. Find a particular solution, z p , to the inhomogeneous

equation by whatever means may be necessary.

Step 3. The sum of the complementary function and this

particular solution is the general solution to the

inhomogeneous equation.

Exercise 12.10. If z c (t) is a general solution to the

complementary equation and z p (t) is a particular solution to

the inhomogeneous equation, show that z c + z p is a solution

to the inhomogeneous equation of Eq. (12.59).



0.2

0.15

0.1

0.05

0



0



0.5



1



1.5



2



2.5



3



3.5



4



t /1 second

FIGURE 12.4 The position of the critically damped oscillator as a

function of time.



Exercise 12.9. Locate the time at which z attains its

maximum value and find the maximum value.



12.3 INHOMOGENEOUS LINEAR

DIFFERENTIAL EQUATIONS: THE

FORCED HARMONIC OSCILLATOR

An inhomogeneous differential equation contains a term

that is not proportional to the unknown function or to any

of its derivatives. An example of a linear inhomogeneous

equation is

d3 z

d2 z

dz

(12.59)

+

f

(t)

+ f 1 (t) = g(t),

2

3

2

dt

dt

dt

where f 3 f 2 , f 1 , and g are some functions of t but do not

depend on z. The term g(t) is the inhomogeneous term. If

an external force F(t) is exerted on a harmonic oscillator,

the oscillator is called a forced harmonic oscillator. If

the external force depends only on the time, the equation

of motion is an inhomogeneous differential equation with

constant coefficients:

f 3 (t)



m



d2 z

= −kz + F(t).

dt 2



(12.60)



12.3.1 Variation of Parameters Method

This is a method for finding a particular solution to a linear

inhomogeneous equation. If the inhomogeneous term is a

power of t, an exponential, a sine, a cosine, or a combination

of these functions, this method can be used. One proceeds

by taking a suitable trial function that contains parameters

(constants whose values need to be determined). This trial

function is substituted into the inhomogeneous equation

and the values of the parameters are found so that the

inhomogeneous equation is satisfied. Table 12.1 gives a list

of suitable trial functions for various inhomogeneous terms.

The trial solution given in this table will not work if the

characteristic equation for the complementary differential

equation has a root equal to the forbidden characteristic

root. If such a root occurs with multiplicity k , multiply the

trial solution by t k to obtain a trial solution that will work.

Example 12.8. Assume that the external force on a forced

harmonic oscillator is given by

F(t) = F0 sin (αt),

where F0 and α are constants. Find the general solution to

the equation of motion.

The equation of motion is

d2 z

F0 sin (αt)

k

.

(12.62)

+ z=

dt 2

m

m

The solution to the complementary equation is the same as

for the unforced harmonic oscillator:

z c = b1 cos(ωt) + b2 sin(ωt).



148



Mathematics for Physical Chemistry











TABLE 12.1 Particular Trial Solutions for the Variation of Parameters

Inhomogeneous term



Trial solution



1



A



tn



A 0 + A1 t + A2 t + · · · + An



e αt



Ae αt



t net



e αt (A



e αt sin (βt )



e αt [A cos (βt ) + B sin (βt )]



α,β



e αt



e αt [A cos (βt ) + B sin (βt )]



α,β



cos (βt )







Forbidden characteristic root

0



0



+ A1 t + A2 t + · · · + An



(ω2 − α 2 )A cos (αt) + ω2 − α 2 B sin (αt)

F0 sin (αt)

,

m



where we have replaced k/m by ω2 . This can be a valid

equation for all values of t only if A = 0 and if

F0

.

m(ω2 − α 2 )



The particular solution is

z p (t) =



α







F0 α

,

m(ω2 − α 2 )

F0 α

vz (0)



.

b2 =

ω

mω(ω2 − α 2 )



Substitution of this into the inhomogeneous equation gives

−α 2 A cos (αt) − α 2 B sin (αt)

k

A cos (αt) + B sin (αt)

+

m

F0 sin (αt)

=

,

m



t n)



vz (0) = b2 ω +



z p = A cos (αt) + B sin (αt).



B=



0

α



Use of Table 12.1 gives the particular solution



=



tn



Exercise 12.11. Find an expression for the initial velocity.



The solution in the previous example is a linear

combination of the natural motion and a motion proportional to the external force. If the frequencies of

these are not very different, a motion such as shown in

Figure 12.5, known as beating, can result. There is a

periodic variation of the amplitude of vibration with a

circular frequency equal to ω − α. You can hear this beating

when a piano is being tuned. There are two or three strings

for each note, and they are tuned separately. Each string

can excite a sympathetic vibration in the other, which acts

as an external force. When the frequencies of two strings

are slightly different, you can hear a pulsation like that in

Figure 12.5, which shows the case that α = 1.100ω.



F0

sin αt .

− α2 )



m(ω2



z



The general solution is

F0

sin αt ,

m(ω2 − α 2 )

(12.63)

where the constants b1 and b2 are to be determined by

the initial conditions. Let us assume that z(0) = 0, so that

b1 = 0.

z(t) = b1 cos(ωt) + b2 sin(ωt) +



t



F0

sin αt ,

− α2 )

dz

F0 α

= b2 ω cos(ωt) +

cos αt .

vz (t) =

dt

m(ω2 − α 2 )

z(t) = b2 sin(ωt) +



m(ω2



The constant b2 would be determined by another initial

condition on vz (0)



FIGURE 12.5 The position of a forced harmonic oscillator as a function

of time for the case α = 1.1ω.



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