Ch17: 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
The Doppler effect
Sound Waves
Sound waves are mechanical longitudinal waves
that propagate in solids, liquids, and gases.
We assume that the surrounding medium is isotropic, i.e., sound propagates with the
same speed for all directions.
The sound wave spreads outward uniformly and the wavefronts are spheres centered at
the point source.
Bulk Modulus
The bulk modulus of the compressed material is defined as:
∆p
B=−
∆V / V
ΔV/V is the fractional change in volume produced by a change in pressure Δp.
The speed of sound
We can show that the speed of sound in a homogeneous isotropic medium with
bulk modulus B and density ρ, is given by the equation:
v=
B
ρ
τ
v=
µ
Medium
Air (20C)
Seawater
Aluminum
Speed (m/s)
343
1522
6420
displacement s(x,t) from its equilibrium position
s( x, t ) = sm cos(kx − ωt )
The constant sm is the displacement
amplitude of the wave (longitudinal).
Δx
s( x, t ) = sm cos(kx − ωt )
Δp from the static value:
∆p( x, t ) = ∆pm sin(kx − ωt )
∆pm = (vρω ) S m
Note: The displacement and the pressure variation have a phase difference of
90˚. As a result, when one parameter has a maximum the other has a minimum
and vice versa.
Amplitude
Oscillating term
phase
s( x, t ) = sm cos(kx − ωt )
Displacement
Angular wave
number
Angular
frequency
∆p( x, t ) = ∆pm sin(kx − ωt )
Pressure variation
Pressure Amplitude
Examples
The pressure in a traveling sound wave is given by the equation:
∆p = 1.5( Pa ) sin[π (0.9 x − 315t )]
Find the pressure amplitude, frequency, wave length and the speed of the wave.
What is the displacement amplitude sm for in air if ρ = 1.2 kg/m3.
What is the expression of the displacement sm(x,t)?
∆pm = 1.5Pa
0.9π = k ⇒ k =
2π
λ
2πf = π × 315 ⇒ f =
⇒ λ = 2.22m
v=
ω
k
315
= 157.5Hz
2
⇒ v = 350m / s
∆pm
1.5
∆pm = ( vρω ) sm ⇒ sm =
sm =
= 3.6µm
vρω
350 × 1.2 × 315 × π
sm = 3.6( µm ) cos[π (0.9 x − 315t )]
Interference
Two point sources of sound waves S1 and S2, are in phase
and emit sound waves of the same frequency, same wave
length.
The point P receives waves from: S1 and S2.
The two waves interfere at point P.
At time t the phase of sound wave 1 arriving from S1 at point P is:
φ1 = kL1 − ωt
At time t the phase of sound wave 2 arriving from S2 at point P is:
φ2 = kL2 − ωt
In general, the two waves at P have a phase difference:
φ = φ2 − φ1 = kL2 − ωt − ( kL1 − ωt ) = kL2 − kL1
φ=
2π
λ
L2 − L1
φ=
2π
λ
L2 − L1
The quantity L2 − L1 is known as the ˝path length difference˝ ΔL
between the two waves
Thus
φ=
2π
λ
∆L
where λ is the wavelength of the two waves.
Examples
Two sound waves, from two different sources with the same frequency, 540 Hz
Travel in the same direction at 330m/s. The sources are in phase. What is the
phase difference of the waves at a point that is 4.4 m from one source and 4m
from the other?
L1= 4m
L2= 4.4m
φ = k L2 − L1 =
φ=
2π
λ
2π
L − L1
v 2
f
ω
L2 − L1
2πf
v= =
= λf
2π
k
λ
2πf
2 × π × 540
φ=
L2 − L1 =
(0.4) = 4.12rad
v
330
Constructive Interference
The wave at P resulting from the interference of the two
waves that arrive from S1 and S2. has a maximum
amplitude when then phase difference
φ = k L2 − L1 =
2π
λ
L2 − L1
φ = 2πm
m = 0,1,2.... ⇒
2π
λ
∆L = 2πm
∆L = mλ
∆L = 0, λ ,2λ ,3λ ....
∆L equals to an integer multiple of λ ↔ Constructive interference
Destructive Interference
The wave at P resulting from the interference of the two waves that arrive from S1 and S2.
has a minimum amplitude when then phase difference
φ = π ( 2m + 1)
m = 0,1,2.... ⇒
2π
λ
∆L = π ( 2m + 1)
1
∆L = ( m + )λ
2
λ 3
5
∆L = , λ , λ ....
2 2 2
∆L equals to a half - integer multiple of λ ↔ Destructive interference
T072
Q6.Two speakers S1 and S2 are placed on the y-axis as
shown in figure 1. The speakers are in phase and emit
identical sound waves with a given frequency. An
observer, standing at point A, hears a sound of maximum
intensity. As the observer moves along a straight line
parallel to the y-axis and reaches point B, he hears first
minimum of sound intensity. The frequency of sound
emitted by the speakers is? (speed of sound in air = 343
m/s).
φA = k L2 − L1 =
φB = k L2 − L1 =
φB = π =
2π
λ
2π
λ
2π
λ
L2 − L1 = 0
12 − 62 + 122 =
2π
λ
(1.41) ⇒ λ = 2.82m
(1.41)
v = λf ⇒ f =
v
λ
=
343
= 121Hz
2.82