Chapter 17 - Waves II
In this chapter we will study sound waves
and concentrate on the following topics:
•Speed of sound waves
•Relation between displacement and
pressure amplitude
•Interference of sound waves
•Sound intensity and sound level
•Sources of Musical sounds
•Beats
•The Doppler effect
• Sound waves are mechanical
longitudinal waves that propagate in
solids, liquids, and gases.
• The figure shows an example of a point
source of sound waves.
• We assume that the surrounding
medium is isotropic, that means that
sound propagates with the same
speed in all directions.
• The sound wave spreads outward uniformly and the wave
fronts are spheres centered at the point source.
•The single arrows indicate lines perpendicular to the wave
fronts called “rays” that point along the direction in which
the sound wave propagates.
•The double arrows indicate the motion of the molecules of
the medium in which sound propagates.
The Speed of Sound
• The speed of any mechanical wave, transverse or
longitudinal, depends on both an inertial property
of the medium (to store kinetic energy) and an
elastic property of the medium (to store potential
energy).
• Thus, we can give the speed of a transverse wave
along a stretched string, by writing
where τ (for transverse waves) is the tension in the
string and μ is the string's linear density.
• If the medium is air and the wave is
longitudinal, we can guess that the inertial
property, μ, corresponds to the volume
density, ρ of air.
• As a sound wave passes through air, potential
energy is associated with periodic
compressions and expansions of small volume
elements of the air.
Bulk modulus
• The property that determines the extent to which
an element of a medium changes in volume when
the pressure (force per unit area) on it changes, is
called the bulk modulus B and is defined as
follows:
SI-unit: pascal
NB: The negative sign indicates that the volume
decreases as the pressure increases.
Replace τ with B μ with ρ then:
• Note 1: Bulk modulus B is smaller for more
compressible media. Such media exhibit lower
speed of sound.
• Nota 2: Denser mediums (larger ρ) have a
smaller speed for sound.
Travelling Sound Wave
Displacement of particles because of a
sound wave
• For transverse waves the medium’s particles oscillate
vertical (y-axis) according to the horizontal (x-axis)
motion of the wave.
• The particles’ displacement are then in the form
y(x,t).
• For longitudinal waves, like sound waves, the
particles of the medium, like air, oscillate parallel to
the wave’s direction of motion, but x(x,t) is
confusing.
• Dus eerder in die vorm s(x,t):
• As the wave travels, the air pressure varies at any
position x sinusiodally as given by:
Phase difference = π/2 or 90˚
Δp = pf - pi < 0 → expansion;
Δp > 0 → compression
Interference
• Two point sources S1 and S2 produce sferical
sound waves that are in phase.
• They move past the same point P.
• In the figure the distance L2 , that the wave
travelled from S2 , is longer than the
distance L1 that the wave from S1 travelled.
• The path difference means that the waves
are possibly out of phase when they reach
P.
Interference
• The phase difference, φ at P depends on the
difference in their path lengths: ΔL = │L2 – L1│
• To relate phase difference φ and the difference
in path lengths ΔL, we use the fact that a
phase difference of 2π rad corresponds to a
distance of one wave length.
• We can write the following ratio:
Then:

L

2

L

2

(17-20)
(17-21)
Derive!
Constructive Interference
•
•
•
•
Takes place when φ = 0, 2π, 4π, 6π, ...
Thus for φ = m(2π), for m = 0, 1, 2, ...
Phase difference = integral multiples of 2π
From eq.17-21 follows: m(2π) = (2π)ΔL/λ
m = ΔL/λ = 0, 1, 2, ... (17-23)
• Thus a path length difference ΔL = 0, λ, 2λ, 3λ, ...,
integral multiples of wave length gives
constructive interference!
Destructive Interference
•
•
•
•
Takes place when φ = π, 3π, 5π, 7π, ...
Thus for φ = (2m+1)π, for m = 0, 1, 2, ... (17-24)
Phase difference = odd numbered multiples π
From eq.17-21 follows: (2m + 1)π = (2π)ΔL/λ
ΔL/λ = 0.5, 1.5, 2.5, ...
(17-25)
• Thus path length difference ΔL = 0.5 λ, 1.5λ, 2.5λ,
..., half-integral multiples of wave length gives
destructive interference!
Intensity of a Sound Wave
Consider a wave that is incident normally on a surface
of area A. The wave transports energy. As a result
power P (energy per unit time) passes through A.
We define the wave intensity I as the ratio P / A :
P
2
I
SI units: W/m
A
The intensity I of a sound wave
at a surface is the average rate
per unit area at which energy is
transferred by the wave
through or onto the surface.
Variation of Intensity with Distance
• In certain cases echoes can be ignored and can
be assumed that the sound source is a point
source that produces isotropic sound— that is
sound with the same intensity in all directions.
• The intensity I can be given by:
where 4πr2 is the area of a sphere.
The Eq. indicates that the intensity of sound from
an isotropic point source decreases with the
square of the distance r from the source.
The Decibel Scale
• The displacement amplitude at the human ear ranges
from about 10-5 m for the loudest tolerable sound to
about 10-11 m for the faintest detectable sound, a
ratio of 106.
• The intensity of a sound varies as the square of its
amplitude, so the ratio of intensities at these two
limits of the human auditory system is 1012. Humans
can hear over an enormous range of intensities.
• Instead of speaking of the intensity I of a sound wave,
it is much more convenient to speak of its sound level
β, defined as: Lower than lowest range of human ear = 10-12W/m2
dB = desibel, the
unit of sound level
β increases with 10 dB each time the sound intensity increases
with a factor 10. Standard reference level corresponds to
zero decibels.
Do Sample Problems 17-4 & 17-5
Sound Standing Waves in Pipes
Consider a pipe filled with air that is open at both ends.
Sound waves that have wavelengths that satisfy a particular
relation with the length L of the pipe set up standing waves
that have sustained amplitudes.
Pipes with two open ends
Standing waves have
antinodes (maximum
displacement amplitude)
at both open ends.
Two antinodes are always
separated by a node
(minimum / zero
displacement) and a
distance of λ/2.
Fig. b shows the “fundamental mode” or “first harmonic”
for the pipe.
The length of pipe, L = λ/2, thus λ = 2L and f = v/λ = v/2L
Standing Waves in Tubes Open at Both Ends
The next three standing wave patterns are
2L
shown in fig. a. The wavelength n 
n
where n  1, 2, 3, .... The integer n is
known as the harmonic number.
nv
The corresponding frequencies f n 
.
2L
2L
n 
n
(17-39) Derive!
Standing Waves in Tubes Open at One End
and Closed at the Other
The first four standing wave patterns are
shown in fig. b. They have an antinode at the
open end and a node at the closed end. L = ½(λ/2) = λ/4
24
LL
The wavelength 

.
n n
n  1n/ 2
Only odd numbered harmonics!
Corresponding
Frequencies
nv
fn 
4L
Do Sample Problem 17-5&6
Beats
(17–12)
Beats
(17-43)
Beats
(17-44)
But because ω1 & ω2 are almost equal, ω >> ω’.
Then Eq.17-45 becomes a cosine function with
angular frequency, ω, and amplitude of the term in
brackets’ absolute value (which varies with angular
frequency, ω’).
Amplitude reaches max twice , when cos function =
-1 or +1.
The Doppler effect
• The phenomena where a change in frequency is
detected when a sound source and detector move
relative to each other.
• We will investigate sound waves in air as medium.
• The speed of the source S and the detector D are
used. We assume that S and D move directly
towards or away from each other against speeds
much smaller than the speed of sound in air.
• Assume that the frequency of the source is equal to
f and the frequency that the detector hears is f’.
If the detector D or the source S (or both) move relative
to each other, the relationship between the original
frequency f and the detected frequency f´ is :
Derive!
•
•
•
•
v = the speed of sound in air
vD = the detector’s speed (relative to air)
vS = the speed of the source (relative to air)
The following rule can be used to determine if the plus or
minus signs must be used:
• NB When the relative motion of source and detector is
towards each other, the answer must give a higher
frequency. If the two move away from each other, the
answer must give a lower frequency.
Detector & Source stationary
We start our investigation with
both the detector & source
stationary:
In time t wave fronts move a
distance, vt
Number of waves that pass D
in time t is, vt/λ
Frequency of waves detected
by D:
vt /  v
f 

t

(17-48)
No Doppler effect!
Detector moves, Source stationary
Wave moves in time t a distance
vt right, while D moves a distance
vDt left. D towards Source.
Wave fronts move in the same
time, t, relative to D, with a speed
v + vD and a distance, vt + vDt.
No. of waves past D: (vt  vDt ) / 
Thus detected frequency at D is
(vt  vD t ) /  v  vD
f '

(17-49)
t

From eq.17-48, follows λ=v/f, thus:
v  vD
v  vD (17-50)
f '
 f
v/ f
v
f‘ > f
Detector moves, Source stationary
If D moves away from source, to
right, a distance vDt follows:
Wave fronts move in the same
time, t, relative to D, a distance,
vt - vDt.
Thus detected frequency at D is
v  vD
f ' f
v
f‘ < f
(17-51)
Thus follows:
v  vD
f ' f
v
(17-52)
Source moves, Detector stationary
Detector D, is stationary, and source
S, moves towards D with speed vS.
In time T, wave front W1,
moves a distance vT, while the
source moves a distance vST.
Speed of sound wave relative to
detector D is thus: v - vS
Distance wave travelled in time T,: λ’ = (v - vS )T
Frequency detected by D:
v
f‘ > f
v
v
f ' 

 ' (v  vS )T (v  vS ) / f
v
 f
(17-53)
v  vD
Source moves, Detector stationary
Detector D, is stationary, and source
S, moves away from D with speed vS.
Now the speed of sound wave
relative to detector D: v + vS
Distance wave travelled in time
T,: λ’ = (v + vS )T
Frequency now detected by D:
v
f ' f
v  vS
(17-54)
f‘ < f
Eq.’s 17-53 &17-54 together gives:
v
f ' f
v  vS
(17-55)
Both source & detector move relative
to each other
• What to do if both source & detector move, but
at different speeds relative to each other?
• Choose the direction of the sound waves that
always move from source towards detector, as
positive, +.
• Determine the corresponding directions of the
source and detector with regards to the direction
of the sound waves:
• Easiest way is to use the following formula:
v  vD
f ' f
v  vS
Sound right :+
Doppler effect f '  f v  vD
vS
vD
+
-
vS
vD
-
+
vD
vS
+
vS
-
+
vD
-
v  vS
v  vD
f f
v  vS
f f
v  vD
f f
v  vS
f f
v  vD
f f
v  vS
v  vD
f f
v  vS
For all
situations:
v  vD
f f
v  vS
Do Sample Problems 17-5,17-6 & 17-8