Producing sound waves
Displacement
1.4 Sound
Density
Pressure
Producing sound waves
Speed of sound
Energy and Intensity
Spherical and Plane waves.
Interference of sound waves
•Produced by compression and rarefaction of media (air)
•Sound waves are longitudinal
resulting in displacement in the direction of propagation.
• The displacements result in oscillations in density and pressure.
Frequencies of sound wave
Speed of sound
Speed of sound in
a fluid
infra-sonic
Audible Sound
10
20,000
v=
ultra- sonic
B=−
Bulk modulus
m
ρ=
V
Frequency (Hz)
30
∆P
∆V / V
B
ρ
0.015
Wavelength (m) in air
Density
Similarity to speed of a transverse wave on a string
v=
elastic _ property
int ertial _ property
Speed of sound in air
v=
B
ρ
v=
γP
ρ
γ is a constant that depends on the nature of
the gas γ =7/5 for air.
Why is the speed of
sound higher in Helium
than in air?
Why is the speed of
sound higher in water
than in air?
P - Pressure
ρ - Density
Since P is proportional to the absolute temperature T by the
ideal gas law. PV=nRT
v is dependent on T
v = 331
T
273
(m/s)
1
Energy and Intensity of sound
waves
Find the speed of sound in air at 20o C.
P=
power
energy
time
area A
v = 331
T
273
v = 331
273 + 20
= 343m / s
273
For calculations use v=340 m/s
Sound intensity level
The decibel is a measure of the sound intensity level
⎛ I⎞
β = 10log ⎜ ⎟
⎝ Io ⎠
Io = 10-12 W/m2
Intensity
I=
power P
=
area
A
(units W/m2)
The ear is capable of
distinguishing a wide
range of sound intensities.
decibels (dB)
the threshold of hearing
note - decibel is a logarithmic unit. It covers a wide range
of intensities.
Question
A = 4πr 2
What is the intensity
of sound at a rock
concert? (W/m2)
area of sphere
As sound spreads out uniformly
from a point source
The intensity decreases as 1/r2
⎛ I⎞
β = 10log ⎜ ⎟ = 120
⎝ Io ⎠
⎛ I ⎞ 120
log ⎜ ⎟ =
= 12
⎝ I0 ⎠ 10
I
= 1012
I0
I = 1012 I0 = 1012 ⋅ 10−12 = 1
Spherical and plane waves
I=
P
4πr 2
W/m2
2
Question 1
Suppose you are standing near a loudspeaker that can
is blasting away with 100 W of audio power. How far
away from the speaker should you stand if you want to
hear a sound level of 120 dB. ( assume that the sound
is emitted uniformly in all directions.)
The sound intensity of an ipod earphone can be as
much as 120 dB. How is this possible?
A)
B)
C)
D)
P
P
I= =
A 4πr 2
r=
P
4πI
=
The ipod is very powerful
The area of the earphone is very small
The ipod is a digital device
Rock music can be very loud
100W
= 2.8m
4π(1W / m2 )
The sound intensity of an ipod earphone can be as
much as 120 dB. How is this possible?
The earphone is placed directly in the ear. The intensity
at the earphone is the power divided by a small area.
Question 2
The sound level in a truck is 20 dB greater than the
sound level in a Strarbucks cafe. If the intensity in the
cafe is 10-7 W/m2 the intensity in the truck is_______
W/m2.
A) 20 X10-7
B) 10-9
C)10-5
D) 20
Say the area is about 1cm2.
P = IA = 1w / m2 (10−4 m2 ) = 10−4 W
A small amount of power produces a high intensity.
Interference of sound waves
Noise canceling headphones
Two sound waves superimposed
Constructive Interference
Noise
Wave 1
Destructive Interference
Wave 2
Anti-noise
3
Interference due to path
difference
Interference
δ=0
path difference =δ =r2 – r1
r1
Constructive Interference
Wave 1
A
r2
Wave 2
Superposition of waves
at A shows interference
due to path differences
In phase at x=0
Condition for constructive interference
Condition for destructive interference
where m is any integer
Sum
δ = mλ
1
2
δ = (m + )λ
λ
m = 0 + 1, + 2,….
Interference
λ
δ= 2
Destructive Interference
Interference
δ=λ Constructive Interference
Interference
δ=
3λ
2
Destructive Interference
Interference
δ=2λ
Constructive interference
4
Interference of sound waves
Phase shift due to path differences
Example
An experiment is performed to measure the speed of sound
using by separating the sound from a single source along
two separate paths with different path lengths and
combining them at the detector. For a frequency of 3.0
kHz (assume vsound =340 m/s);
A) What would the smallest path difference be to observe a
minimum in intensity
x
r2 – r1 =
When
r2 –r1 =mλ
When r2 – r1 = (m+½) λ
m is any integer
Constructive Interference
Destructive Interference
340m / s
λ v
=
=
= 5.7cm
2 2f 2(3 x103 s −1 )
B) What would the smallest (non-zero) path difference be
to observe a maximum in intensity.
r2 – r1 =
λ = 11cm
Example 14.6 Path difference for two sources.
r2-r1 =0.13 m
At position P the listener hears the first minimum in
sound intensity. Find the frequency of the oscillation.
vsound =340 m/s
At position P the path difference is equal to λ/2. (first
minimum) destructive interference.
λ
= r2 − r1 = 0.13m
2
λ = 2(0.13) = 0.26m
v 340m / s
f= =
= 1.31x103 Hz
λ
0.26m
5