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)
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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
2π
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
2π
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
2τ
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
=
2π
2π
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.
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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 =
2π
=
1
2π
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.
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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.