Honors Classical Physics I
PHY141
Lecture 32
Sound Waves
Please set your clicker to channel 21
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Lecture 32
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Real Loudspeakers
• Bosch 36W column loudspeaker polar pattern
• Monsoon Flat Panel
speaker:
(5 dB grid)
– 400 Hz:
– 1 kHz:
– 3 kHz:
– 7 kHz:
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Lecture 32
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Example
Your ears are sensitive to differences in pitch, but they
are not very sensitive to differences in intensity. You are
not capable of detecting a difference in sound intensity
level of less than 1 dB.
• By what factor does the sound intensity increase if the
sound intensity level β increases from 60 dB to 61 dB ?
I2
I1
   2  1  (10 dB) log10  (10 dB) log10
 1 dB
I0
I0
I2
I 2 I1
 (10 dB) log10
 (10 dB) log10  1 dB
I0 I0
I1
I2
  100.1  ...
I1
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Example
A 3 db Increase in sound level corresponds to:
+3 dB = 10 dB log(Inew/Iold)  Inew/Iold = 103/10 = 100.3 = 2.0
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Lecture 32
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A 10 dB increase in the sound level
corresponds to an increase in intensity of
factor of …
Rank
Responses
1
2
3
4
5
6
Other
β(I) ≡ (10 dB) log10(I /I0)
with I0≡10–12 W/m2
Δβ(I) ≡ β(If) – β(Ii) = (10 dB) log10(If /Ii)
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Standing Sound Waves
• In open or closed air pipes sound waves will form
standing waves when the air is rhythmically excited at
the appropriate frequency… This is the basis for all
wind-instruments: organs, flutes, etc.
• Note that the velocity of sound in air is fixed (v = 344
m/s for normal conditions), thus the product of fλ = v is
fixed, and f and λ are NOT independent!
• As for standing waves on a string under tension, we
expect the wavelength λ of the standing wave, and the
length L of the pipe, to be similarly related.
– At an OPEN END of the pipe, the pressure is closely equal to the
ambient atmospheric pressure: i.e. there must be a PRESSURE
NODE
– NO such situation occurs for a CLOSED END; in fact there the
PRESSURE has an ANTI-NODE (maximum amplitude)
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Wind Pipes
Considering the above statements, we arrive at the following picture
(where we depict an OPEN-ENDED pipe, and graph the standing
PRESSURE waves Δp(x,t) on top:
L
λ2/2
λ1/2
λ4/2
– A harmonic series appears (we show the 1st, 2nd, and 4th harmonic), governed
by the statement that a “whole number of half-wavelengths must fit in
length L of the pipe”:
• i.e.: L = n(λn/2), with: n=1,2,3,… or: λn = 2L/n = λ1/n or: fn = v/λn = n v/(2L) = n f1
the DISPLACEMENT graph looks different: the place where pressure
has a NODE, displacement has an ANTI-NODE and vice versa, because
there the motions of the molecules (displacing themselves to keep local
pressure constant) is most violent…
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Stopped Wind Pipes
Consider now a STOPPED pipe: a pipe with one OPEN
end, and one CLOSED END (for example the clarinet),
and graph the standing PRESSURE waves Δp(x,t) on top:
L
λ1/4
λ2/4
λ4/4
Careful inspection shows that a different harmonic
series appears (we show the 1st, 3rd, and 7th harmonic):
• i.e.: L = (2n – 1) λn/4 , with: n=1,2,3,… (Note: (2n – 1) is odd!)
• equivalently: L = nodd λn/4 , with: n=1,3,5,… (ODD harmonics only)
• or: λn = 4L/nodd = λ1/nodd or: fn = v/λn = nodd v/(4L) = nodd f1
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A one-end-open, one-end-closed organ pipe
is 0.34 m long; the ground tone has a
frequency of … (Hz)
Rank
1
2
3
4
5
6
Responses
vsound = 344 m/s
L
Other
λ1/4
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Lecture 32
λ2/4
60
λ4/4
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Resonance
Large displacement/pressure waves will occur when the exciting force is
acting in sync with the “NATURAL FREQUENCY” of the pipe (or other
type of instrument). Absent damping, the displacements can become
uncomfortably large!
– Resonance is used in many instruments to enhance particular
(over)tones (e.g. bass reflex tube, violin/guitar body case)
Example: A 0.40 m long, one-side-closed organ pipe is in exact resonance
with a 0.50 m guitar string; both vibrate at the fundamental tone.
– Calculate vstring: Lpipe=λ1pipe/4 and Lstring=λ1string/2
 fpipe= vair/(λ1pipe=4Lpipe) = fstring= vstring/(λ1string=2Lstring)
 vstring=vair 2Lstring/(4Lpipe) = 0.625×344 m/s = 215 m/s
– Calculate the λ and f of the sound waves in the air:
 λ1= 4Lpipe= 1.60 m, f1= vair/λ1=344/1.60 = 215 Hz
– Calculate the λ and f of the standing wave on the guitar string:
 λ1=2Lstring=1.0 m; check: f1= vstring/λ1=215/1.00=215 Hz
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Interference
Interference is the destructive or constructive addition of displacements by
waves arriving from two or more sources at a set of spatial locations.
– Interference is a consequence of the SUPERPOSITION PRINCIPLE, which says that
disturbances caused by individual waves at any given point simple ADD – this is a
consequence of the linear character of the wave equation
– Thus, a trough from one wave may coincide with an equally high peak of another, with
the result there is no dis-placement at all: destructive interference. Such a “dead
spot” will persist if the two waves have exactly the same frequency.
– If the waves have a slightly different frequency, “beat waves” may occur:
 (kx  1t)  (kx  2t) 
 (kx  1t)  (kx  2t) 
y (x,t)  A cos(kx  1t)  A cos(kx  2t)  2 A cos 
cos



2
2




      

1
 2Acos kx  1 2 t  cos  1 2 t   2Acos(kx t) cos 1 t   2Acos t  cos(kx t)
2
2


2
2








– This is a traveling wave, with a frequency equal to the average frequency of the initial
waves, and an amplitude which is modulated (i.e. varies in time) with a much smaller
frequency equal to half the difference of the original frequencies.
The INTENSITY is proportional to the amplitude squared, and thus the beat
frequency we hear is simply equal to the (absolute value of) the difference in
original frequencies:
1

I  cos 2   t 
2

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1  cos( t) 
double-angle formula 2

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λ
maxima
(twice the amplitude)
Interference by SameFrequency Waves
minima
(dead spots)
y
A
x
B
•P
Interference of two
synchronous equalfrequency, equalamplitude sound sources, ignoring reflections
from walls, floor, ceiling,
etc…
Movie…
Maximum in P: |AP – BP| = nλ, n=0,1,2,…;
Minimum in P: |AP – BP| = (n+ ½)λ
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Doppler Effect: fL=fS(v+vL)/(v+vS)
Doppler effect: frequency of received sound depends on the relative
motion of the source or receiver with respect to the medium (e.g. air):
λ’
λ’<λ
f’>f
Doppler Formula:
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λ’’
λ
v  vL
fL  fS
v  vS
Lecture 32
+ve for L
approaching S
λ’’>λ
f’’<f
+ve for S
moving away
from L
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Doppler Effect - Derivation
Doppler effect: frequency of received sound depends on the relative
motion of the source or receiver with respect to the medium (e.g. air):
– E.g. moving towards the source of the sound wave, my ears will “catch” more
pressure variations per second than if I stay still or move away.
• Thus, the frequency fL I perceive depends on my velocity vL with respect to the air.
• Similarly, the motion of the sound source with respect to the air vS affects the
wavelength λ of the air waves.
For the simple case where all motions are along the x-direction:
– Assume a sound wave in –ve x-direction; the speed of sound is v = 343 m/s
λ
– LISTENER who has velocity vL in +ve x-direction:
vL
v
vL  (  v )
v  vL
relative speed
fL




x
– SOURCE with velocity vS in +ve x-direction: the source travels a distance vST
per period, so that the effective wavelength is increased by that amount:
vS
v λ
wavelength
  S vS T 
v  vS
v
v
 S 
fS
fS
fS
Combining: Doppler Formula: f L 
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x
v  vL
Lecture 32

v  vL
 fS
v  vS
+ve for L
approaching S
+ve for S
moving away
from L
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Example
v  vL
fL  fS
v  vS
+ve for L approaching S
+ve for S moving away from L
– Note, that if vL = vS then fL = fS (e.g. when the wind blows from source to
receiver, nothing changes…
a bat emits a high-pitched “chirp” at 80 kHz (ultra-sound)
when it approached a fixed wall with velocity vBat=10 m/s
– calculate the frequency of the reflected chirps the bat receives…
v
 f reflected from wall
fincident on wall  f Bat
v  vBat
v  vBat
v v  vBat
f received by Bat  f reflected from wall
 f Bat
v
v  vBat
v
v  vBat
354
 f Bat
 80 kHz
 84.8 kHz
v  vBat
334
– Note that the SIGN of the velocities is CRUCIAL!
– the problem is even more complex when the wall is a MOVING INSECT!
This is an example of SONAR
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A strong wind is blowing from a stationary
source towards a stationary listener …
A. The received tone is lower than the emitted tone …
B. The received tone is equal to the emitted tone …
C. The received tone is higher than the emitted tone …
33%
fL  fS
v  vL
v  vS
33%
+ve for L
approaching S
+ve for S
moving away
from L
A.
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33%
Lecture 32
B.
C.
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