PHYS1110H, 2011 Fall.
Shijie Zhong
Wave Motion
Definitions:
Mechanical waves: disturbances of material medium, such as air, water,
a string. Electromagnetic waves: light waves, radio waves, X rays, that
do not need a material medium. We deal with mechanical waves here.
A wave moves energy from a place to another place, but not matter.
Longitudinal and transverse waves: If the matter oscillates back and
forth in the same direction as the direction of wave propagation, the wave is
called longitudinal wave (top figure above). Sound wave is an example. If
the matter oscillates up and down in a direction perpendicular to the
direction of wave propagation, the wave is called a transverse wave (bottom
figure above). The wave on a guitar string is a transverse wave. Some waves
have both longitudinal and transverse waves.
Wave shape and amplitude: Wave shape is also called waveforms. An
isolated disturbance is a pulse (figure below). A continuous wave is an
ongoing periodic disturbance (bottom figure above). The wave train is an
intermediate case between a pulse and a continuous wave. The amplitude of
a wave is the maximum value of wave disturbance.
1 Wavelength, period, and frequency: A continuous wave repeats itself
in space and time. Take a snapshot of a continuous wave, wavelength λ is
the distance over which the wave repeats itself (see the figure above). Take a
point in space where a continuous wave travels, the time for a complete
oscillation of a wave is called period T (figure above). The frequency f is the
number of wave cycles per unit time (e.g., per second), and is the inverse of
period T.
Wavespeed: It is the speed at which a wave travels in a material medium.
For example, sound speed is ~340 m/s, seismic wave speed is several km per
second. Wavespeed is often determined by the properties of media. For
example, sound speed is higher for damp air than dry air.
v=! T =!f
(1)
A mathematical description of sinusoidal w av e:
For a continuous wave with a single frequency, the following equation
describes the wave completely as a function of time t and position x:
2 y(x,t) = A cos(kx " #t) ,
!
where y is the disturbance of the wave (e.g., pressure in sound wave or
displacement of string for wave on a guitar string), A is the amplitude of the
wave, k is wavenumber k=2π/λ, and ω is angular frequency ω=2πf. Notice
v=
!
(2)
" #
= .
T k
(3)
For (2), the waveform can be obtained for a fixed time t=t0, i.e., a snapshot
in time, y(x,t0 ) = A cos(kx " #t0 ) with only independent variable of position
x. For a fixed location or x0, y(x0 ,t) = A cos(kx0 " #t) describes the
disturbance as a function of time t (see the figure above).
!
!
!
!
Equation (2) describes a wave traveling in +x direction. This can be seen in
the following way. Suppose that we look at the wave crest at x0 of a
snapshot at t0 (e.g., the solid curve in the figure above), y(x0, t0)=A or
kx0 " #t0 = 0 . As the wave travels or time t increases to t0+dt (e.g., the
dashed curve for the snapshot at this time), the wave crest’s position x would
have to increase from x0 to x0+dx, where dx = " / kdt = vdt , to maintain
kx0 " #t0 = 0 or y(x0, t0)=A (i.e., wave crest). With the increase in x, this
means that the wave crest travels in +x direction.
x t
Note that equation 2 can also be!written as y(x, t) = A cos[2! ( ! )] .
" T
For a wave travels in –x direction, y(x,t) = A cos(kx + "t) .
Interference:
!
Two wave trains from opposite
directions meet. The total disturbance is the
sum of that for the two waves. This is called superposition principle.
This principle applies for any number of waves. For N numbers of waves
3 with different wavenumber and angular frequency, when they travel and
meet in a space (e.g., different sound waves in a room), the total disturbance
at a location and time is given by
N
y(x,t) = $ Ai cos(ki x " # i t) ,
(4)
i=1
where i is the index for i-th wave.
!
Constructiv e interference: when two waves with the same wavelength
meet, if they have the same phase, then they coincide and lead to
constructive interference.
Destructiv e interference: when two waves with the same wavelength
meet, if their phases are offset by π, these two waves cancel each other and
lead to destructive interference.
The left figure shows two pulses
meet, destructively interfere, and
then go their separate ways.
Standing w av es: Two sinusoidal traveling waves of same wavelength and
the same amplitude, traveling in the opposite directions on a string with
length L, reflecting at the ends of string, make a standing wave.
4 y(x, t) = A cos(kx ! ! t) ! A cos(kx + ! t) = 2Asin kx sin ! t .
(5)
For a given location of the string, x, the amplitude and frequency of
oscillation are 2Asinkx and ω, respectively. Nodes are where the amplitude
of the oscillation on the string is zero. For example, at x=0 and L, i.e., the
fixed ends of the string. Particularly, kL=nπ, where n is an integer, 1, 2, 3, ...
2!
L = n!
"n
or
!n =
2L
n .
(6)
n=1 is the fundamental mode, and n=2 is the first overtone. Including nodes
at x=0 and L, the total number of nodes is equal to n+1 (see the figure
below).
0
L
Standing waves are not limited to vibrations on a string. Waves in confined
regions including water waves in a pond or sounds in musical instruments all
form standing waves.
W av e equation
In physics, some differential equations are very important. For example, we
discuss the 2nd order ODE for oscillatory motion: d2x/dt2+ω2x=0. We now
derive another differential equation that describes wave motions.
5 Let us consider waves on a string or vibration of a spring which is a
transverse wave. The string’s two ends are fixed, and the string is under
some tensional force T (see the figure below). Consider a small segment dx,
from x to x+dx of the string. The linear density of the string is ρ. Let y(x,t)
be the displacement or vibration of the string at position x and time t. We
will derive a differential equation that describes y(x,t).
For the small segment dx, its horizontal motion is negligible, hence
(7)
T (x + dx)cos(! (x + dx)) ! T (x)cos(! (x)) = 0 ,
where θ is the angle of the segment dx to its equilibrium position (i.e.,
small), and θ is position-dependent, and
θ=tan(θ)=dy/dx.
(8)
For small θ, cos(! (x + dx)) ! 1, cos(! (x)) ! 1 , equation (7) becomes
T (x + dx) = T (x) .
(9)
Now for the vertical motion of the segment, the Newton’s 2nd law:
d2y
T (x + dx)sin(! (x + dx)) ! T (x)sin(! (x)) = " dx 2 ,
(10)
dt
Considering that sinθ=θ for small angle, the left hand side of (10) is,
d!
T (x)[sin(! (x + dx)) ! sin(! (x))] = T (x)[! (x + dx) ! ! (x)] = T (x) dx .
dx
Substituting this back to (10) leads to
d!
d2y
T
=" 2 ,
or considering (8),
dx
dt
d2y
d2y
T 2 =! 2 .
(11)
dx
dt
Defining v = T / ! , equation (11) becomes
6 d2y 1 d2y
=
,
dx 2 v 2 dt 2
or
!2 y 1 !2 y
=
!x 2 v 2 !t 2
(12)
Equation (12) is called the wave equation and the second form uses partial
derivatives (Note: A partial derivative is that the derivative is taken with
respect to a variable, while holding the others as constant).
The solution to (12) describes waves, as we show below. It can be shown
that y(x, t) = f (x ! vt) is a solution, where f is an arbitrary function.
Defining z=x-vt, using the chain rule, and considering !z !x = 1 and
!z !t = "v , we have
!y !f !z !f
=
= ,
!x !z !x !z
!y !f !z
!f
=
= "v ,
!t !z !t
!z
(13)
The left hand side of (12) is:
!2 y ! !y
! !y !z ! !y
! !f
!2 f
= ( )= ( ) = ( )= ( )= 2 .
!x 2 !x !x !z !x !x !z !x !z !z
!z
The right hand side of (12) is:
(14)
1 !2 y 1 ! !y
1 ! !y !z
1 ! !y
! !f
!2 f
=
(
)
=
(
)
=
"
(
)
=
(
)
=
.
(15)
v 2 !t 2 v 2 !t !t
v 2 !z !t !t
v !z !t
!z !z
!z 2
This proves that y(x, t) = f (x ! vt) is a solution of equation (12). In deriving
(14) and (15), we considered (13). Likewise, one can prove
y(x, t) = f (x + vt) is also a solution.
Notice that f can be an arbitrary function and describes a waveform (pulse or
sinusoidal). It can be proved that y(x, t) = f (x ! vt) , and y(x, t) = f (x + vt)
describe waves with waveform f that travel at wave speed of v to positive x
and negative x directions, respectively. The proof is similar to how we
proved for sinusoidal wave traveling in page 3. However, we can consider a
arbitrary waveform f (see the figure below).
7 Suppose that at t=0, the wave pulse is centered at x=0 as y(x, 0) = f (x) . At a
later time t=t1, the waveform is given by y(x, t1 ) = f (x ! vt1 ) . Define
x ' = x ! vt1 and y(x, t1 ) = f (x ') , which represents the same waveform at t=0
but with the wave shifting by a distance d=vt1 in the +x direction. This
proves that y(x, t1 ) = f (x ! vt1 ) represents a wave traveling at a speed of v
towards +x direction.
Remarks:
1) Wave speed on a string v = T / ! is controlled by tension and density of
the string, i.e., how tight the string is setup and what materials the string is
made of.
2) Sinusoidal wave given in equation (2) satisfies wave equation (12).
2"
#
2"
y(x, t) = A cos(kx ! ! t) = A cos[ (x ! t)] = A cos[ (x ! vt)] ,
#
T
#
which has the form of y(x, t) = f (x ! vt) .
3) The general solution to (12) can be written as
y(x, t) = f (x ! vt) + g(x + vt) ,
where g is another function that may differ from f.
(16)
Sound w av e and its equation
Sound waves are longitudinal waves. As sounds travel in a tube, the air
pressure in the tube is perturbed. The perturbed pressure travels, thus making
the sound. Consider a tube fill with air with unperturbed density ρ0 and
pressure P0 (see the figure below). Suppose that a sound wave is excited and
traveling in the tube. Our goal is to develop a differential equation that
describes the sound wave.
Consider a small segment of tube Δx, from x to x+Δx and the tube has a
cross-sectional area A. Suppose that air particles at x is displaced
horizontally (i.e., in the same direction as x) by ξ(x), and at x+Δx, by
8 ξ(x+Δx). The differential displacement between the two ends of this small
segment, ξ(x+Δx) - ξ(x), leads to volume change of this small segment of
the tube, and hence density and pressure changes.
!V = V " V0 = A[! (x + !x) " ! (x)] = A[! (x) +
d!
d!
d!
!x " ! (x)] = A!x
= V0
dx
dx
dx
V ! V0 d!
,
(17)
=
V0
dx
The relative volume change (i.e., the left hand side of equation 17) can be
related to relative change in air density as following.
V ! V0 m ! ! m !0
! ! !0
"!
,
(18)
=
=!
=!
V0
m !0
!
!
where m is the mass of the gas in the given volume. Combining (17) and (18)
leads to
d!
""
.
=!
dx
"
(19)
For an ideal gas, the gas law is PV =nRT. At constant temperature T, change
in volume V leads to change in pressure P, such that PV =Pm/ρ=nRT or P/ρ
is a constant C.
!P !!
,
(20)
=
P0
!0
where Δ, as usual, represents the change. However, the density change
associated with the pressure change in sound waves is NOT at constant
temperature or NOT isothermal process, rather the process is adiabatic (i.e.,
no exchange of heat with surrounding media). Equation (20) for isothermal
process needs to be revised by multiplying a constant factor γ,
P = C ! , or
!P = C!! ,
or
9 !P
!"
=!
P0
"0
or
!P !"
.
=
! P0 "0
(21)
γ is a constant of order 1 and is dependent on the type of gas or air, because
it is the ratio of specific heat at constant temperature to specific heat at
constant pressure (You will learn more in studies of thermodynamics in the
future). For air or diatomic gases, γ is 7/5, but for monatomic gases like
helium, it is 5/3.
Combining (19) and (21), and considering ! ! !0 , we have
d!
"P
,
=!
dx
" P0
or !P = "! P0
d"
.
dx
(22)
Finally, consider Newton’s 2nd law for the small segment. The mass of this
segment, m = ρA Δx. The net force acting on the segment results from the
pressure difference between the two ends of the segment (see the figure
above),
Fnet = !AP(x + "x) + AP(x) = A[!"P(x + "x) + "P(x)] ,
(23)
Newton’s 2nd law:
d 2"
A[!"P(x + "x) + "P(x)] = ! A"x 2 .
dt
(24)
Use Taylor expansion, !P(x + !x) = !P(x) +
d("P)
d 2"
!
=! 2 .
dx
dt
d[!P(x)]
!x , and equation (24)
dx
(25)
Substituting ΔP from (22) to (25) leads to
d 2"
d 2"
! P0 2 = # 2 ,
dx
dt
or
!2!
" !2!
, or
=
!x 2 # P0 !t 2
!2! 1 !2!
=
,
!x 2 v 2 !t 2
(26)
10 ! P0
is the wave speed for the sound wave, and (26) has the
"
same form as (12) that describes wave motions. Likewise, the solution for
(26) is the same, ! (x, t) = f (x ! vt) for sound waves traveling to +x direction.
where v =
Remarks:
1) Sound wave speed is ~343 m/s, but depending on temperature and
pressure, this may vary by +-10 m/s. Sound wave speed is much slower than
the speed light. This is why we always see a lighting first before we hear the
thunder.
2) Humans can hear sounds with frequency ranges from f=20 Hz to 20,000
Hz. Given the sound speed of ~343 m/s, it is straightforward to compute the
wavelength ranges using λ=v/f.
3) For sound wave, ! (x, t) = A cos(kx ! " t) , the amplitude A determines how
loud the sounds are, while the frequency ω determines the “pitch” of the
tone.
11