Energy and Intensity of sound
waves
Sound II.
power
P=
energy
time
area A
Energy and Intensity
Interference of sound waves
Standing waves
Complex sound waves
Intensity
I=
power P
=
area
A
(units W/m2)
Sound intensity level
The ear is capable of
distinguishing a wide
range of sound intensities.
⎛ I⎞
β = 10log ⎜ ⎟
⎝ Io ⎠
Io = 10-12 W/m2
decibels (dB)
the threshold of hearing
⎛ I ⎞
β = 10log ⎜ ⎟
⎝ Io ⎠
β
I
= 1010
Io
decibel is a logarithmic unit. It covers a wide range
of intensities.
Question
Question
The sound intensity of an ipod earphone can be as
much as 120 dB. How is this possible?
What is the intensity
of sound at a rock
concert? (W/m2)
⎛ I⎞
β = 10log ⎜ ⎟ = 120
⎝ Io ⎠
A)
B)
C)
D)
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
⎛ I ⎞ 120
log ⎜ ⎟ =
= 12
⎝ I0 ⎠ 10
I
= 1012
I0
I = 1012 I0 = 1012 ⋅ 10 −12 = 1
W/m2
1
Spherical and plane waves
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.
A = 4πr 2
area of sphere
For a point source the intensity
decreases as 1/r2
Say the area is about 1cm2.
I=
P = IA = 1w / m2 (10−4 m2 ) = 10−4 W
P
4πr 2
A small amount of power produces a high intensity.
P = power of source
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.)
I=
r=
Interference of sound waves
Two sound waves superimposed
Constructive Interference
P
P
=
A 4πr 2
P
4πI
=
Destructive Interference
100W
= 2.8m
4π(1W / m2 )
Interference due to path
difference
Noise canceling headphones
path difference =δ =r2 – r1
r1
Noise
Wave 1
Wave 1
A
r2
Wave 2
Anti-noise
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
δ = mλ
1
2
δ = (m + )λ
m = 0 + 1, + 2,….
2
Interference
δ=0
Constructive Interference
Interference
λ
δ= 2
Destructive Interference
Sum
λ
Interference
Interference of sound waves
Phase shift due to path differences
δ=λ Constructive Interference
x
When
r2 –r1 =mλ
When r2 – r1 = (m+½) λ
m is any integer
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
r2 – r 1 =
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 – r 1 =
λ = 11cm
Constructive Interference
Destructive Interference
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
3
Standing Wave
• A standing wave is formed by reflections
back and forth at the boundaries of a
media.
• The standing wave does not carry energy
but serves to store energy.
• The standing wave stores energy of waves
with specific wavelengths.
Standing Waves
Standing waves (waves on a
string)
Standing waves in air columns.
Standing Wave with 3 nodes
– At different times.
Standing wave on a string
Nodes
The Standing wave
doesn’t
“move” it just stands in
one place
Anti-nodes
Fixed end
Standing Waves
Standing Wave Conditions
One node at each end with
additional nodes only at specific
positions
L
• A standing wave is generated by superposition
of two waves with the same frequency and
wavelength traveling in opposite directions.
Simulation of a standing wave.
n=1
n=2
Distance between nodes = λ/2
A string with length L can support
standing waves of only at certain
wavelengths.
L=
http://www.walter-fendt.de/ph14e/stwaverefl.htm
n=3
λ
n
2
where n=1, 2, 3… integer values
n=1 called the fundamental
n=2 called the second harmonic etc.
4
Standing wave frequencies and
wavelengths
L=
n=1
λ
n
2
λn =
fn =
n=2
Example
Find the fundamental and second harmonics of a steel
wire fixed at both ends. The speed of the wave in the
string is v=200 m/s
2L
n
v
v
=
n
λ 2L
λ1 = 2L
f1 =
v
2L
λ2 = L
f2 =
v
L
=
200
= 100Hz
2(1)
=2f1 =200 Hz
L
where n = 1, 2, 3......
n=3
Standing waves in air columns
Fundamental Frequency
2 ends open
2 ends closed
N
L
L=
N
λ1
2
A
A
L=
λ1= 2L
λ1
2
λ1= 2L
v
F1=
2L
F1=
v
2L
v is the speed of sound in air
one end open
one end closed
N
Cylinder open at both ends
Harmonics
A
L=
λ1
4
λ1=
4L
F1=
v
4L
fn = n f1
F1 lower by a
factor of 2
Cylinder open at one end closed at
one end - Harmonics
n = 1, 2, 3, 4 ....... All harmonics
Summary
For a cylinder with the same length
5f1
Frequency
4f1
3f1
2f1
f1
0
fn =nf1
n= 1, 3, 5, 7 .... Only odd harmonics
v
n
2L
fn = f1n
7f1
5f1
fn =
n=1,2,3...
3f1
all harmonics
f1
n=1, 3, 5.......
only odd harmonics
fn =
both ends
open/ closed
v
n = f1n
4L
0
one open
one closed
5
Example
A cylinder 5.0 cm in length is closed at one end and
open at the other end. Find the frequency of the third harmonic
of the standing wave in the column. vair =340 m/s
L
λ1 = 4L
4L
3
λ 3=
f3 = 3f1 =
3v
4L
=
Musical Instruments
String Instruments
v
v
f1= λ = 4L
1
3(340)m / s
= 5.1x103 Hz
4(0.05)m
Musical Instruments
Wind instruments
The sound is produced by vibrating air and the
frequency is enhanced by resonance in the
air column
Frequency due to standing waves on the string.
The body of the instrument acts as a resonator to move
air to amplify the sound.
Complex waves
• In general sound waves are a combination
of different frequencies.
• The superposition of waves with different
frequencies gives rise to the characteristic
quality (timbre) of the sound.
• The different frequencies can be
determined by mathematical procedure
called a Fourier Transform.
Complex waves consist of different frequency
components , i.e. harmonics.
displacement
Time
relative
amplitude
Frequency
6