Download 1.2 Modeling of Harmonic Waves

1.2
Modeling of Harmonic Waves
1.2.1
Properties of Harmonic Waves
Objective 1: Identify the wavelength, frequency and period for a harmonic wave.
Objective 2: Use the equation for energy of an electromagnetic wave to calculate the energy,
frequency and wavelength of an electromagnetic wave.
The periodic change of a variable can be described as a simple harmonic and can be modeled by sine or
cosine functions. Figure 1.4 is a sine wave illustrating periodic variation of the wave variable for both
transverse and longitudinal type waves.
y
λ
A
Wave
propagation
or speed, v
x
Figure 1.4 Schematic of sine wave illustrating the typical wave characteristics of amplitude (A),
wavelength (λ), and wave propagation speed (v).
If Figure 1.4 represents a transverse wave, where there is a particle displacement perpendicular to the
wave propagation, the wave can be characterized by the following:
A
is the amplitude or the maximum distance the particle is displaced in the y direction.
λ
is the wavelength and is equivalent to 1 complete cycle.
f
is the frequency and is equal to the number of complete cycles (equivalent to 1 wavelength) the
particle completes per second. The wavelength, frequency and wave propagation speed (v) are
related by,
· 1
w is the angular frequency and is equal to 2πf.
τ
is the period of the wave and is defined as the time required for the particle to complete 1 cycle (1
wavelength) and is equal to 1/f. The period is sometimes represented as T.
If this wave represents the movement of a particle, the particle speed would differ from the wave
propagation speed where the particle path follows along the sine curve.
The characteristics (A, λ, f, w and τ) used to describe the harmonic motion of a particle can also be used
to describe the motion of water in water waves, the changing air pressure in sound waves, and the
varying electric field in electromagnetic waves.
Example: Determine the wavelength of an electromagnetic wave (c = 3 x 108 m/s) that has a frequency
of 1015 s-1. Give the answer in units of nanometers. (1 m = 10-9 nm). Identify the region of the
electromagnetic spectrum that corresponds to this wavelength.
Solution:
3 10 10 300 10 1 Electromagnetic radiation with wavelength of 300 nm corresponds to the visible spectrum (visible light
has wavelengths between 300 and 700 nm). Note: A distinction is made between visible light and light,
where the latter refers to all electromagnetic radiation.
Example: Repeat the previous example for an electromagnetic wave with frequency 1019 s-1. What kind
of electromagnetic radiation do we have now?
Solution:
3 10 10 0.03 10 1 Electromagnetic radiation with wavelength of 0.03 nm corresponds to X-ray radiation (X-ray radiation
has wavelengths between 0.01 – 1 nm).
Example: Radio waves have the largest wavelengths of all electromagnetic radiation. Determine the
frequency of a radio wave with wavelength equal to 3,000 m.
Solution:
3 10 10 3000 Note: A wave that has a large wavelength also has a small frequency within the electromagnetic
spectrum.
1.2.2
Mathematical Representation of Waves as a function of time
Objective 1: Recall the wave equations for a harmonic wave as a function of time.
Objective 2: Graph a harmonic wave as a function of time and understand how the wave parameters
affect the wave.
Objective 3: Understand the effect of angular frequency on the wavelength.
The following sine and cosine functions are used to model the periodic harmonic (oscillating) motion of
a particle as a function of time,
sin cos where y represents the motion or displacement of the particle about the equilibrium position (y = 0) as a
function of time.
Note that if A is held constant and the angular frequency is increased by 2, the wavelength is cut in half.
Example: Derive the expression that relates the angular frequency (w) to the wavelength (λ).
Solution:
2!
Therefore, if w is increased by 2, the wavelength will be cut in half. This makes sense, because both the
frequency and wavelength are related to the energy. Increasing the wave frequency will increase the
wave energy, and this is demonstrated by a decrease in the wavelength. This energy-frequencywavelength relationship is a fundamental wave relationship that is applicable to any kind of wave.
Example: Model 2 electromagnetic waves with the function E = Eo cos (wt), where Eo is the electric field
amplitude. Assume Eo1 = Eo2 = 1 V/cm, and that w1 = 5 s-1 and w2 = 2w1 = 10s-1. Compare the
wavelength for the 2 waves.
Solution:
Increasing the
angular frequency
(w) by 2 resultsin a
decrease in the
wavelength of the
second wave by
exactly 50%.
1.2.3
Mathematical Representation of Waves as a function of distance and time
Objective 1: Recall the wave equation for a traveling harmonic wave as a function of x and t (w and
k). (This equation will be used to derive Schrödinger’s Equation)
Objective 2: Understand how the wave variables w and k affect the wave characteristics (λ and f).
Objective 3: Use the harmonic wave equation to graph a traveling harmonic wave for several values
of t and to evaluate the difference in phases between the waves.
Consider 2 frames of reference, one stationary (y) and one moving at a speed c (y’) with respect to the
stationary frame (in this case c is an arbitrary velocity and not equal to the speed of light).
y
&
Y’
v
x
Figure 1.5 Propagation of a sine wave with respect to 2 reference frames, one stationary frame and one
propagating at speed c relative to the stationary frame. The time lag between the two waves
is equal to t – x/v.
The expression y = sin wt can be written as a function of both t and x. The expression for t can be
related to the speed v and distance x. At any time t, the second frame of reference can be related to the
first by,
", $ sin % & '
The particle traveling in the y’ reference frame would have traveled for a time x/v, so we need to
subtract this amount of time in order to arrive at the stationary point on the stationary reference frame
y. If the above function is graphed for two values of t, we can observe the difference in phases between
the two waves.
The equation ", $ sin % & ' is commonly expressed as a function of other variables as
indicated below.
(
", $ sin 2! ) & +
* where
.
, /
3
0
12
0
", $ sin" & 4$
where
.
5
5
6
12
4
12
/
4 7 89798 :
5
5;9 <=><:?:7> >:
,
3
>= ;9 :9 @A9=
Plotting the equation ", $ sin" & 4$ as a function of distance only (with t = 0) results in a
typical sine wave. Adding a second curve for a time t, will result in a shift of the sine wave by t/τ cycles
in the negative x direction. This occurs because we are subtracting kx from wt. The equation
", $ sin" & 4$ is for a harmonic wave traveling in the positive x direction. For a wave
traveling in the negative x direction, the equation becomes ", $ sin" B 4$.
1.2.4
Differential Wave Equation
Objective 1: Recall the 2nd order differential wave equation for a harmonic wave as a function of t and x.
Objective 2: Recall the general solutions to the differential wave equation.
Objective 3: Apply differential calculus to prove if a function is a solution to the 2nd order differential
wave equation.
The equations derived in section 1.2.3 describe a wave propagating at a speed c through a medium. The
value of c differs from the particle velocity, v, through the medium. The particle velocity can be
determined by taking the partial derivative of y with respect to t at constant x.
", $ sin" & 4$
C
cos" & 4$
C
The acceleration of the particle can be determined by taking the second partial derivative with respect
to t.
:
C1
& 1 sin " & 4$
C 1
The first and second order partial derivatives of y with respect to x at constant t are obtained as follows.
C
&4 cos " & 4$
C
Dividing the equation for
DE F
DG E
C1
&4 1 sin" & 4$
C 1
DE F
by the equation for D( E results in the following expression.
C 1 H
C 1
C 1 H
C 1
& 1 sin " & 4$
1
1
&4 1 sin" & 4$
41
since, w = 2πf,k = 2π/λ and v = fλ.
Rearranging the above expression results in the following second order differential equation,
C1
C1
1
C 1
C 1
This 2nd order differential equation is called the wave equation for a simple harmonic oscillator. This
equation describes the motion of a wave at speed c along x at time t. Specifically, y is the displacement
in the y direction for a string wave. The variable y represents the changes in pressure for a sound wave
traveling through air or a different medium. For an electromagnetic wave, y is the value of the electric
or magnetic field.
The general solutions to this equation for a “free particle” that travels as a monochromatic (constant w)
harmonic wave in a straight line at constant speed, c, and constant A (undamped), can include any of
the following:
cos " & 4$
sin" & 4$
exp 7" & 4$
exp &i" & 4$
The general solution to a 2nd order differential equation must include 2 constants. Since the equation
DE F
DG E
1 D( E is linear, then the following statement is true:
DE F
if functions f and g are general solutions to the wave equation, then the function (f + g) is also a general
solution, and therefore the following are also general solutions,
>" & 4$ B L sin" & 4$
and
M exp 7" & 4$ B N exp &7" & 4$
where A, B, C and D are constants. Note that these 2 solutions are equivalent, which can be proven
using trigonometric formulae.
The constants in the above equations can be determined by applying boundary conditions. For 2
undetermined constants, 2 boundary conditions are required.
Example: Prove that the general solution y = 9< & i" & 4$ is a solution of the differential wave
equation for a harmonic wave.
Solution:
In order to prove that the expression is a general solution of the differential wave equation, we need to
determine the expressions for
get the same expression.
DE F
D( E
DE F
and DG E , and substitute these into the differential wave equation to
Using the following definition for the derivative of an exponential function,
determined.
DE F
D( E
DE F
and DG E can be
O
O(
"9 P $ "9 P $
OP
O(
The partial derivative of y with respect to x (t constant) is calculated as follows,
y = 9< & i" & 4$
y = 9<"&7 B 74$
C
74 C
C1
7 1 4 1 &4 1 C 1
The partial derivative of y with respect to t (x constant) is calculated as follows,
C
&7 C
C1
7 1 1 & 1 C 1
Substituting these expressions into the wave equation and simplifying results in,
C 1 H
C 1
C 1 H
C 1
and this is equal to,
"2!$1
& 1 1
"$1 1
&4 1 41
"2!H$1
C1
C1
1
C 1
C 1
which is the differential wave equation and therefore y = 9< & i" & 4$ is a solution of this
equation.
Homework
1.
Expand Table 1.1 by adding a column for frequency (f) and energy (E) as shown below. Calculate
the values of f and E for the low and high range wavelength values using the relation E = hc/λ
and E = hf where c = 3 x 108 m/s and h = 6.63 x 10-34 J s. Use the conversion 1 eV = 1.6 x 10-19 J
to calculate values of Energy in units of eV. Note that the unit of s-1 is equivalent to 1 Hertz (Hz).
Wavelength*
> 10 cm
Radio Waves
10 cm – 0.01 cm
Micro Waves
700 nm – 0.01 cm
Infrared Waves
400 – 700 nm
Visible Waves
400 nm – 10 nm
Ultraviolet Waves
0.01 nm – 10 nm
X-Rays
< 0.01 nm
Gamma Waves
* Approximate values
2.
Frequency, s-1
-
Energy, eV
-
Graph y = A sin(wt) for a range of t between 0 and 0.2 seconds in increments of 0.01 for the
following:
(a) A = 0.01 m, f = 5 s-1, where w = 2πf
(b) A = 0.01 m, double the value of w
(c) Compare the wavelengths for parts (a) and (b)
3.
Graph y = A sin(wt – kx) for a range of x between 0 and 0.012 m in increments of 0.0003 for the
following conditions (A = 0.01 m, f = 5 s-1, λ = 0.005 m):
(a) t = 0
(b) t = 0.25 seconds (graph on the same graph as part (a))
(c) Determine the difference in phase between the two graphs.
4.
5.
6.
Graph y = A sin(wt – kx) for a range of x between 0 and 2 m in increments of 0.1 for the
following conditions:
(a) A = 1 m, f = 1 s-1, λ = 1 m, t = 0
(b) A = 1 m, f = 1 s-1, λ = 1 m, t = 0.5
(c) What is the value for τ for the wave in (a) and (b)?
(d) What is the value for t/τ for the wave in (a) and (b)?
(e) What happens if you add wave (a) to wave (b) and plot the resulting wave?
Prove that the general solution y = exp &i" & 4$ is a solution of the differential wave
equation for a harmonic wave.
Prove that the general solution y = A sin " & 4$ is a solution of the differential wave
equation for a harmonic wave.
References
1
2.
3.
4.
5.
D.A.B. Miller, Quantum Mechanics for Scientists and Engineers, Cambridge University Press, New York, 2008.
A. Beiser, Concepts of Modern Physics, McGraw Hill, New York, 2003.
nd
J.R. Taylor, C.D. Zafiratos, M.A. Dubson, Modern Physics for Scientists and Engineers (2 Ed.), Prentice Hall,
New Jersey, 2004.
F.W. Sears, Zemansky, Young, Addison Wesley Education Publishers, 1991.
G. Parker, Introductory Semiconductor Physics, Institute of Physics Publishing, London, 2004.