2.1
A mathematical model is a description
of a real-world system using
mathematical language and ideas.
Exponential Growth
Differential equations are fundamental to modern science and engineering. Many of
the known laws of physics and chemistry are expressed as differential equations, and
differential equation models are used extensively in biology, economics, and other
social sciences.
The most common use of differential equations in science is to model dynamical
systems, i.e. systems that change in time according to some fixed rule. For such a
system, the variable under consideration (such as mass M, population P, or electric
charge Q) is a function of time t. Thus, a typical differential equation might be written
dM
t3 M2
dt
instead of the more familiar
y0 x 3 y 2 .
The Exponential Growth Equation
Perhaps the most common differential equation in science is the equation for exponential growth. It can be used to model a variety of phenomena in the natural and
social sciences, including unrestricted population growth, nuclear chain reactions,
continuously compounded interest, and macroeconomic growth.
The Exponential Growth Equation
The exponential growth equation is the differential equation
dy
k y,
dt
where k is a positive constant. Its solutions have the form
y y0 e kt ,
where y0 y (0) is the initial value of y.
The formula y y0 e kt represents exponential growth, as shown in Figure 1. The
constant k is called the growth constant, and has units of inverse time (e.g. number per
second).
a Figure 1: Exponential growth.
EXAMPLE 1
Solve the following initial value problem.
dy
3y,
dt
SOLUTION
y (0) 4.
According to the formulas above, the solution is y ( t ) 4 e 3t .
EXPONENTIAL GROWTH
2
EXAMPLE 2
Solve the following initial value problem.
dy
3y,
dt
y (2) 8.
SOLUTION This time the initial value is at t 2 instead of t 0, so we have some work to do.
We know that y has the form
y ( t ) y0 e 3t
for some constant y0 . Plugging in y (2) 8 gives
8 y0 e 6 ,
so y0 8 e −6 . Then
y ( t ) 8 e −6 e 3t 8 e 3t−6 .
Population Growth
Exponential growth is often used to model the growth of populations of organisms in a
resource-rich environment. Here “resource-rich” means that there is plenty of food
and other resources necessary for the population to grow indefinitely.
The relative growth rate of a population P ( t ) is the quantity
1 dP
.
P dt
This represents the rate of growth of P relative to the size of the population. For
example, if a bacteria culture has a relative growth rate of r 3/hour, it means
that each individual bacterium has an average of 3 offspring per hour (not counting
grandchildren).
Assuming a population P mantains a constant relative growth rate, its growth will
satisfy the differential equation
dP
rP.
dt
Such a population grows exponentially with time:
r P P0 e rt
a Figure 2: Exponential population growth.
where P0 is the population at time t 0 (see Figure 2).
EXAMPLE 3
During a biology experiment, a certain culture of cells has a constant relative growth rate of
0.04/minute. If there are 5,000 cells at the beginning of the experiment, how large will the
culture be one hour later?
SOLUTION
The population P ( t ) is given by the equation
P ( t ) 5,000 e 0.04t .
We are measuring t in minutes, so one hour later corresponds to t 60. Since
P (60) 5,000 e 0.04 (60) ≈ 55,115.9,
the population after one hour will be approximately 55,000 cells.
EXPONENTIAL GROWTH
3
EXAMPLE 4
A population of bacteria initially consists of 20,000 cells. Twenty minutes later, the population
has grown to 50,000 cells. How quickly is the population increasing at that time?
SOLUTION
Assuming constant relative growth rate, the population P ( t ) is given by the
equation
P ( t ) 20,000 e rt .
(1)
for some constant r. Plugging in P (20) 50,000 gives the equation
50,000 20,000 e r (20) .
To solve the equation
20,000 e 20r 50,000
for r, we first divide by 20,000 to get
e 20r 2.5,
and then we take the natural logarithm of
both sides, which yields
20r ln (2.5) .
We can now divide through by 20 to
obtain the value of r.
and solving for r yields
ln 2.5
≈ 0.045815.
20
To find the rate of increase at t 20 min, we must compute P 0 (20) . Taking the derivative of
equation (1) gives
P 0 ( t ) 20,000 r e rt
r so
P 0 (20) 20,000 (0.045815) e (0.045815)(20) ≈ 2290.73.
Thus the population is increasing at a rate of roughly 2,300 cells/min.
Describing the Growth Rate
For an exponential growing quantity y y0 e kt , there are a few alternatives to using
the growth constant k to describe the rate of exponential growth. The simplest is to use
the time constant τ, which is the reciprocal of the growth constant:
τ 1
.
k
Using the time constant, the formula for exponential growth can be written
y y0 e t/τ .
The time constant τ has units of time, and represents the amount of time that it takes
for the variable y to be multiplied by a factor of e. For this reason, τ is also known as
the e-folding time.
Another way to describe the rate of exponential growth is to use the doubling
time Td , which is the amount of time that it takes for y to double in size. The doubling
time satisfies the equation
e kTd 2
and hence
ln 2
(ln 2) τ.
k
Since ln 2 ≈ 0.6931, the doubling time is approximately 69% of the e-folding time.
Td EXAMPLE 5
A small tumor is growing exponentially with a doubling time of 15 days. If the tumor currently
has 1.3 million cells, how quickly is the number of cells increasing?
SOLUTION
Since the tumor is growing exponentially, the number of cells N is given by the
formula
N ( t ) N0 e kt ,
EXPONENTIAL GROWTH
4
where N0 1.3 million cells. Since the doubling time is 15 days, we know that
e 15k 2.
Taking the natural logarithm of both sides and then dividing through by 15 gives
k ≈ 0.046210/day.
To find N 0 (0) , we take the derivative of the formula for N to get
N 0 ( t ) N0 k e kt .
Plugging in t 0 gives
N 0 (0) N0 k (1.3 million cells)(0.046210/day) 60 thousand cells/day.
EXERCISES
1. Solve the following initial value problem:
dy
3y,
dt
y (ln 2) 40.
2. At the beginning of an experiment, a culture of E. coli contains 15,000 cells. Two
hours later, the population has grown to a size of 80,000 cells. Assuming exponential
growth, what is the time constant τ?
3. During a nuclear chain reaction, the number N of free neutrons in a sample of
fissile material obeys the differential equation
dN
kN,
dt
where k a constant known as the neutron multiplication factor.
(a) Suppose the sample initially contains 100 free neutrons. If the neutron multiplication factor is 7.0/µs, how many free neutrons will there be 2.0 µs later?
(b) How quickly will the number of free neutrons be increasing at that time?
4. The population of the United States is currently 320 million, and is increasing at a
rate of 2.4 million/year. Assuming exponential growth, what is the doubling time
of the population?
5. A cell culture is growing exponentially with a doubling time of 3.00 hours. If
there are 11,500 cells initially, how long will it take for the cell culture to grow to
30,000 cells?
2.2
Exponential Decay
When the growth constant k is negative, the function
y y0 e kt
does not actually grow. Instead, the value of y approaches zero as t → ∞. This is
known as exponential decay.
When discussing exponential decay, it is common in applications to write k −r,
where r is a positive constant known as the decay constant. In this case, the differential
equation takes the following form.
The Exponential Decay Equation
The exponential decay equation is the differential equation
dy
−r y
dt
This is just the exponential growth
equation with −r substituted in for k.
where r is a positive constant. Its solutions have the form
y y0 e −rt
where y0 y (0) is the initial value of y.
Figure 1 shows the graph of a typical exponential decay. As with exponential
growth, there are two alternatives to using the decay constant r when describing
exponential decay:
a Figure 1: Exponential decay.
1. The time constant (or e-folding time) of y is the quantity τ 1/r, and represents
the amount of time that it takes for the value of y to be divided by e.
2. The half-life of y is the amount of time that it takes for the value of y to be cut in
half.
As we shall see, exponential decay can be used to model such diverse phenomena as
radioactive decay, electric circuits, and chemical reactions.
Radioactive Decay
See the EPA website for a list of
radioisotopes commonly used in industry.
A substance whose atoms are inherently unstable is called radioactive. For such a
substance, a certain fixed proportion of the atoms decay during each time interval. If N
is the number of atoms in a sample of the substance, then N will satisfy the differential
equation
dN
−rN.
dt
Here r is the decay constant, which represents the rate at which individual atoms tend
to decay. For example, if r 0.003/hour, it means that about 0.3% of the atoms will
decay each hour.
The number of atoms in a radioactive sample decays exponentially with time:
N N0 e −rt ,
a Figure 2: Exponential decay of a
radioactive sample.
where N0 is the number of atoms int the sample at time t 0 (see Figure 2). Equivalently,
the total mass M of atoms in a sample will decay exponentially:
M M0 e −rt .
EXPONENTIAL DECAY
2
EXAMPLE 1
Cesium-137 has a half-life of approximately 30.17 years. If a 0.300-mole sample of 137 Cs is left
in a storage closet, how much 137 Cs will be left after four years?
SOLUTION
The amount N ( t ) of 137 Cs will obey an equation of the form
N ( t ) 0.30 e −kt ,
where k is a constant. Since the half-life is 30.17 years, we know that
1
.
2
e −k (30.17) Solving for k gives k ≈ 0.022975/year. Then
N (4) 0.300 e − (0.022975)(4) 0.273659,
so there should be 0.274 moles left after four years.
EXAMPLE 2
A radiochemist prepares a cobalt sample containing 0.100 moles of 58 Co. According to readings
from a Geiger counter, the atoms of 58 Co in the sample appear to be decaying at a rate of
6.79 × 10−7 moles/min. Based on this information, what is the half-life of 58 Co?
SOLUTION
The amount N ( t ) of 137 Cs will obey an equation of the form
N ( t ) 0.100 e −kt ,
where k is a constant. Then
N 0 ( t ) −0.100 k e −kt .
N 0 (0) is negative since N is decreasing.
We are given that N 0 (0) −6.79 × 10−7 moles/min, which gives us the equation
−6.79 × 10−7 −0.100 k.
Solving for k gives k 6.79 × 10−6 /min. The half-life is the value of t for which
e −kt 1
.
2
Plugging in k and solving for t yields a half life of 102,084 minutes, which is about 70.9 days.
RC Circuits
Figure 3 shows a simple kind of electric circuit known as an RC circuit. This circuit has
two components:
capacitor
• A resistor is any circuit component—such as a light bulb—that resists the flow of
electric charge. Applying voltage to a resistor will force current through it, with
the amount of current given by Ohm’s law
resistor
I a Figure 3: An RC circuit.
V
.
R
(1)
Here V is the applied voltage, I is the resulting current, and R is a constant called
the resistance of the resistor.
EXPONENTIAL DECAY
3
• A capacitor is a circuit component that stores electric charge. A charged capacitor
can supply voltage to a circuit, with the amount of voltage given by the equation
V Q
.
C
(2)
Here Q is the charge stored in the capacitor and C is a constant called the
capacitance of the capacitor.
All of the circuit-related quantities
discussed here have their own SI units:
• Electric charge is measured in
coulombs (C).
In an RC circuit, the voltage produced by a capacitor is applied directly across a resistor.
Substituting equation (2) into equation (1) yields a formula for the resulting current:
I • Voltage is measured in volts (V).
• Electric current is measured in
amps (A), where 1 A 1 C/sec.
Q
.
RC
(3)
This current represents the flow of charge out of the capacitor, with
• Resistance is measured in
ohms (Ω), where 1 Ω 1 V/A.
dQ
−I.
dt
• Capacitance is measure in
farads (F), where 1 F 1 C/V.
Substituting equation (3) into this equation yields the differential equation
dQ
Q
−
.
dt
RC
Thus the charge Q will decay exponentially, with decay constant
Equivalently, the time constant for the RC
circuit is
τ RC.
r 1
RC
In the case where the resistor is a light bulb, the result is that the bulb lights up at first,
but becomes dimmer and dimmer over time as the capacitor discharges.
EXAMPLE 3
A 0.25 F capacitor holding a charge of 2.0 C is attached to a 1.6 Ω resistor. How long will it
take for the capacitor to expend 1.5 C of its initial charge?
SOLUTION
The charge on the capacitor will decay exponentially according to the formula
Q Q 0 e −rt ,
where Q 0 2.0 C and
Since seconds are the SI unit of time, the
units for r must be number per second.
r 1
1
2.5/sec.
RC
(1.6 Ω)(0.25 F)
If the capacitor expends 1.5 C of its charge, it will have 0.5 C left. Plugging this into the
formula for Q gives
0.5 2.0 e − (2.5) t ,
and solving for t yields t 0.5545. Thus the capacitor will expend 1.5 C of charge in
approximately 0.55 seconds.
EXPONENTIAL DECAY
4
EXERCISES
1. A sample of an unknown radioactive isotope initially weighs 5.00 g. One year later
the mass has decreased to 4.27 g.
(a) How quickly is the mass of the isotope decreasing at that time?
(b) What is the half life of the isotope?
2. During a certain chemical reaction, the concentration [C4 H9 Cl] of butyl chloride
obeys the rate equation
d[C4 H9 Cl]
−r [C4 H9 Cl],
dt
where r 0.1223/sec is the rate constant for the reaction. How long will it take for
this reaction to consume 90% of the initial butyl chloride?
3. A capacitor with a capacitance of 5.0 F holds an initial charge of 350 C. The
capacitor is attached to a resistor with a resistance of 9.0 Ω.
(a) How quickly will the charge held by the capacitor initially decrease?
(b) How quickly will the charge be decreasing after 20 seconds?
4. An LR circuit consists of a resistor attached to an electrical component called an
inductor, which supplies voltage to the resistor according to the formula
V −L
dI
.
dt
Here L is a constant called the inductance of the inductor.
(a) Combine the equation above with Ohm’s law to obtain a differential equation
for the current I ( t ) that involves the constants L and R.
(b) The current I ( t ) in an LR circuit decays exponentially. Find a formula for the
decay constant in terms of L and R.
5. For a planetary atmosphere of ideal gas of uniform temperature T, the atmospheric
The first of these equations is the molar
form of the ideal gas law, while the
second is the equation for hydrostatic
pressure.
pressure P ( h ) and density ρ ( h ) at a height h above the ground are related by the
equations
dP
−ρ g,
P ρRT
and
dh
where R is the specific gas constant and g is the acceleration due to gravity.
(a) Combine the given equations to obtain a single differential equation for P
involving the constants g, R, and T.
(b) The atmospheric pressure in such an atmosphere varies with height according
to the formula P ( h ) P0 e −rh , where P0 is the pressure at ground level. Find a
formula for the decay constant r in terms of g, R, and T.
2.3
Harmonic Oscillation
So far, we have used differential equations to describe functions that grow or decay
over time. The next most common behavior for a function is to oscillate, meaning that
it increases and decreases in a repeating pattern. The simplest kind of oscillation is
harmonic oscillation.
The square root in the formula for ω
comes from taking the derivative twice.
Specifically, if
The Harmonic Oscillator Equation
The harmonic oscillator equation is the differential equation
d2 y
−r y,
dt 2
y A cos ( ωt + φ ) ,
then
y 0 −ωA sin ( ωt + φ ) ,
where r is a positive constant. Its solutions have the form
and
y A cos ( ωt + φ ) ,
y 00 −ω2 A cos ( ωt + φ ) ,
so
y 00 −ω 2 y.
where ω √
r, and A ≥ 0 and −π < φ ≤ π are constants.
Figure 1 shows a typical harmonic oscillation. The constant A is the amplitude of
the oscillation, and is equal to the maximum value of y. The constant φ is the phase
constant, which is measured in radians, and the constant ω is the angular frequency,
which is measured in radians per unit time.
There are a few common alternatives to angular frequency for measuring the rate of
oscillation:
a
Figure 1: Harmonic oscillation.
1. The period T is the amount of time it takes for the oscillation to complete one
full cycle. It is related to the angular frequency of a harmonic oscillation by the
formula
T The SI unit for frequency is the hertz (Hz),
which represents one cycle per second.
2π
ω
2. The frequency f is the number of cycles per unit time. The frequency is related
to the period T and the angular frequency ω for a harmonic oscillation by the
formulas
f 1
ω
T
2π
EXAMPLE 1
A harmonic oscillator satisfies the differential equation
d2 y
dt 2
−0.34 y.
What is the period of oscillation?
SOLUTION
The angular frequency is ω √
2π
0.34 ≈ 0.5831, so T ≈ 10.78.
ω
HARMONIC OSCILLATION
2
Cartesian Form
There is a nice alternate form for harmonic oscillations that involves two terms added
together.
Harmonic Oscillations (Cartesian Form)
The general solution to the differential equation
d2 y
−r y,
dt 2
can also be written as
y C 1 cos ( ωt ) + C 2 sin ( ωt ) ,
where ω √
r and C 1 and C 2 are arbitrary constants.
Given a harmonic oscillator
y A cos ( ωt + φ ) ,
we can convert it to Cartesian form by applying the angle sum formula for cosine:
Here we are using the formula
cos ( a + b ) cos a cos b − sin a sin b.
y A cos φ cos ( ωt ) − A sin φ sin ( ωt ) .
Thus y C 1 cos ( ωt ) + C2 sin ( ωt ) , where
C 1 A cos φ
C 2 −A sin φ
and
Conversely, given the values for C 1 and C 2 , we can find the amplitude A and phase
constant φ using the formulas
A q
C1 2 + C2 2 ,
cos φ C1
,
A
sin φ −
C2
.
A
The advantage of Cartesian form is that it vastly simplifies the task of solving initial
value and boundary value problems.
EXAMPLE 2
Solve the following initial value problem.
d2 y
dt 2
SOLUTION
−25y,
y (0) 3,
y 0 (0) 10.
We know that
y ( t ) C 1 cos (5t ) + C 2 sin (5t )
for some constants C1 and C 2 , which means that
y 0 ( t ) −5C 1 sin (2t ) + 5C 2 cos (2t ) .
In general, if
y C1 cos ( ωt ) + C2 sin ( ωt )
y ( t ) 3 cos (5t ) + 2 sin (5t ) .
then
y (0) C 1
Plugging y (0) 3 into the first equation and y 0 (0) 10 into the second equation yields C 1 3
and C 2 2, so the solution is
and
y 0 (0) ωC2 .
HARMONIC OSCILLATION
3
EXAMPLE 3
Solve the following boundary value problem.
d2 y
−4y,
dt 2
SOLUTION
y (0) 1,
y
π
8
√
3 2
We know that
y ( t ) C 1 cos (2t ) + C2 sin (2t )
for some constants C1 and C 2 . The two boundary values yield the equations
1 C1
and
√
π
π
+ C 2 sin
.
3 2 C 1 cos
4
4
Plugging C1 1 into the second equation and solving for C 2 gives C 2 5, and hence
y ( t ) cos (2t ) + 5 sin (2t ) .
EXERCISES
1. A harmonic oscillator y ( t ) has phase constant π/3. Given that y (0) 5, what is
the amplitude?
2. Express the harmonic oscillator y ( t ) 10 cos 6t +
π
in Cartesian form.
4
3. Find the amplitude and phase constant for each of the following harmonic oscillators.
(a)
(b)
(c)
(d)
y (t )
y (t )
y (t )
y (t )
cos (3t ) + sin (3t )
3 cos (2t ) + 4 sin (2t )
sin (5t )
√
√
− 2 cos t + 6 sin t
4. Solve the following initial value problem:
d2 y
−5y,
dt 2
y (0) 3,
y 0 (0) 10.
5. Solve the following boundary value problem:
d2 y
−y,
dt 2
y (0) 4,
y
2π
3
1.
6. Solve the following boundary value problem:
d2 y
−4y,
dt 2
π
y −
6
4,
π
y
6
16.
2.4
Harmonic Oscillator Models
In this section we give three important examples from physics of harmonic oscillator
models. Such models are ubiquitous in physics, but are also used in chemistry, biology,
and the social sciences to model oscillatory behavior.
Mass-Spring Systems
spring
block
Consider the simple mass-spring system shown in Figure 1, which consists of a block
with mass m attached to spring whose other end is fixed. According to Hooke’s Law,
the force that the spring exerts on the block is given by the equation
F −kx.
frictionless table
a Figure 1: A simple mass-spring system.
Here k is the spring constant, and x is the horizontal position of the block, with x 0
being the rest position.
Newton’s second law (F ma) can be written as
F m
We are neglecting friction in this
computation. Adding a frictional force to
the model would make this an example
of damped harmonic oscillations.
d2 x
,
dt 2
where x ( t ) is the horizontal position of the block. Assuming the spring is the only
horizontal force affecting the block, this gives us the differential equation
−kx m
d2 x
,
dt 2
which we can rewrite as
d2 x
k
−
x.
2
m
dt
This is a form of the harmonic oscillator equation, with angular frequency
An object whose position oscillates
harmonically is said to undergo
harmonic motion.
r
ω k
m
EXAMPLE 1
A 3.0 kg mass is attached to a spring with a spring constant of 4.0 kg/sec2 . The spring is
stretched 0.80 m from its rest position and then the mass is released. What is the speed of the
mass 1.0 sec later?
SOLUTION
The mass will undergo harmonic motion with an angular frequency of
r
ω k
m
r
4.0
1.1547 rad/sec.
3.0
The position x ( t ) of the mass obeys a formula of the form
x ( t ) C 1 cos ( ωt ) + C2 sin ( ωt ) ,
for some constants C1 and C 2 . Taking the derivative gives
x 0 ( t ) −ωC1 sin ( ωt ) + ωC 2 cos ( ωt )
We are given that x (0) 0.80 and x 0 (0) 0, and plugging these in gives C1 0.80 and C 2 0,
so the speed of the mass at t 1.0 sec is
x 0 (1.0) − (1.1547)(0.80) sin (1.1547) −0.84 m/sec.
HARMONIC OSCILLATOR MODELS
2
LC Circuits
capacitor
inductor
a Figure 2: An LC circuit.
Figure 2 shows a simple kind of electric circuit known as an LC circuit. This circuit
consists of a capacitor connected to a circuit component inductor, which is essentially
just a coil of wire. Unlike a resistor, which always resists the flow of current, an inductor
tends to oppose changes to the flow of electric current. That is, it’s difficult to start
pushing current through an inductor, but once the current gets going, it’s difficult to
make it stop.
The voltage drop V across an inductor is given by the formula
V L
The SI unit of inductance is the
henry (H), where 1 H 1 V · sec/A.
dI
dt
Here L is a constant called the inductance of the inductor. Note that the inductor has
positive voltage drop (like a resistor) when the current is increasing, but when the
current is decreasing the voltage drop is negative, meaning that the inductor actually
pulls current through it.
Combining the formula above with the equation V Q/C for the capacitor yields
L
Q
dI
.
dt
C
As in an RC circuit, the electric current I is that same as the rate at which the capacitor
is discharging, so
I −
dQ
.
dt
Substituting this into the previous equation yields the differential equation
−L
Q
d2Q
− .
C
dt 2
which we can rewrite as
d2Q
1
−
Q.
2
LC
dt
This equation describes simple harmonic oscillation with angular frequency
ω √
1
LC
Thus the charge held in the capacitor oscillates according to the formula
Q A cos ( ωt + φ ) ,
The capacitor is like the “spring” in an LC
circuit (with spring constant 1/C), while
the inductor is the “mass”.
where A is the amplitude of the oscillations (measured in coulombs, the SI unit of
charge), and φ is the initial phase of the circuit.
Roughly speaking, if we assume that the capacitor begins charged, then the capacitor
begins by discharging through the inductor, slowly at first but picking up speed as the
inductor lets more current through. Once the capacitor is fully discharged, the inductor
continues pushing current through the circuit, which drains even more charge from
the capacitor, leaving it with a negative total charge. The capacitor then reverses the
flow of current to regain the lost charge, but the same thing happens again, with the
inductor continuing to push current through in the reverse direction until the capacitor
is back to its initial charged state. The cycle thus continues indefinitely.
HARMONIC OSCILLATOR MODELS
3
EXAMPLE 2
An LC circuit consists of a capacitor with a capacitance of 0.016 F and an inductor with an
inductance of 0.10 H. The capacitor starts with an initial charge of 0.12 C, and the initial
current is zero. What is the magnitude of the current in the circuit 0.10 seconds later?
SOLUTION
The charge stored on the capacitor will oscillate harmonically with
ω √
1
LC
√
1
(0.016)(0.10)
25 rad/sec.
Then
Q ( t ) C 1 cos (25t ) + C 2 sin (25t )
for some constants C1 and C 2 , with derivative
Q 0 ( t ) −25C 1 sin (25t ) + 25C2 cos (25t ) .
We are given that Q (0) 0.12 and Q 0 (0) 0, and plugging these in gives C1 0.12 and
C2 0. Then
Q 0 (0.10) − (25)(0.12) sin (2.5) −1.7954.
Thus the current at time t 0.10 sec is approximately 1.8 A.
Pendulums
pivot
A pendulum consists of a mass suspended from a rod that swings from a fixed pivot
point, as shown in Figure 3. If θ ( t ) denotes the angle of the string from the vertical,
then θ obeys the differential equation
rod
g sin θ
d2 θ
−
,
2
L
dt
mass
a Figure 3: A pendulum.
a Figure 4: Anharmonic motion of a
pendulum with initial conditions
θ (0) 0.99π and θ0 (0) 0.
where g is the acceleration due to gravity (usually 9.8 m/sec2 ), and L is the length of
the string.
Unfortunately, equation (1) is not an instance of the harmonic oscillator equation,
because the right side involves sin θ instead of θ. This means that a pendulum is actually
an anharmonic oscillator. For example, Figure 4 shows the noticeably anharmonic
motion of a pendulum that starts from rest at θ (0) 0.99π.
Though the motion of a pendulum is anharmonic, we can make a harmonic
approximation for the motion in the case where θ isn’t too large. This depends on the
linear approximation
sin θ ≈ θ,
which is quite accurate when θ is close to zero, as shown in Figure 5. Indeed, as long as
θ stays between −14◦ and 14◦ , this approximation is accurate to within 1%. Replacing
sin θ with θ in equation (1) gives us the approximate differential equation
g
d2 θ
≈ −
θ.
L
dt 2
This equation describes approximate harmonic motion, with angular frequency
r
ω ≈
a Figure 5: The graphs of y sin θ and
y θ nearly coincide for θ close to zero.
(1)
g
L
HARMONIC OSCILLATOR MODELS
4
EXAMPLE 3
A swinging pendulum with a length of 2.0 m has an initial position of θ (0) 0.10 rad and an
initial angular velocity of θ0 (0) −0.12 rad/sec. What will the position of the pendulum be
0.80 sec later?
SOLUTION
The pendulum will move approximately harmonically with angular velocity
r
ω ≈
g
L
r
9.8
2.2136 rad/sec2 .
2.0
Then
θ ( t ) ≈ C1 cos ( ωt ) + C 2 sin ( ωt )
and
θ0 ( t ) ≈ −ωC1 sin ( ωt ) + ωC 2 cos ( ωt )
for some constants C1 and C 2 . Plugging in the initial conditions gives C 1 0.10 rad and
C2 −0.05421 rad. Then
θ (0.50) ≈ (0.10) cos (1.77088) − (0.05421) sin (1.77088) −0.0730036,
so the pendulum will be at an angle of approximately −0.073 rad.
HARMONIC OSCILLATOR MODELS
5
EXERCISES
1. A pendulum with a length of 0.30 m starts from rest at an angle of 0.18 rad. How
quickly will the angle of the pendulum be changing 0.20 sec later?
2. A 2.0 µF capacitor is connected to an inductor. If the resulting system oscillates
with a frequency of 3.0 kHz, what is the inductance of the inductor?
3. A 3.5 kg mass has been attached to a spring with a spring constant of 24 kg/sec2 . If
the mass is oscillating with an amplitude of 1.6 m, what is the maximum speed of
the mass during the oscillation?
4. A charged capacitor with a capacitance of 3.0 F is attached to an inductor with an
inductance of 0.20 H. The initial current in the circuit is zero, but after 1.0 sec the
current has increased to 8.0 A. What was the initial charge on the capacitor?
5. When two masses m 1 and m 2 are connected by a spring, the length L of the spring
obeys the differential equation
m1 m2 d 2 L
−kL,
m 1 + m 2 dt 2
where k is the spring constant.
(a) Suppose a 3.0 kg mass is attached to a 5.0 kg mass by a spring with a spring
constant of 12 kg/sec2 . What is the period of the resulting oscillations?
(b) In a carbon monoxide (CO) molecule, the bond between the carbon and oxygen
atoms can be modeled as a spring with spring constant 1.13 × 1039 u/sec2 .
Given that the carbon atom has a mass of 12.0 u and the oxygen atom has a
mass of 16.0 u, at what frequency will the molecule tend to vibrate?
6. A pendulum with a length of 20.0 cm is started from rest at an angle of 45.0◦ .
(a) What period does the harmonic approximation predict for this pendulum?
Assume that the acceleration due to gravity is 9.81 m/sec2 .
(b) The pendulum is measured to have an actual period of 0.933 seconds. What was
the percentage error in the period predicted by the harmonic approximation?