PES 2130 Fall 2014, Spendier
Lecture 13/Page 1
Lecture today: Chapter 17 Waves-2
1) Standing sound waves (Sources of Musical Sound)
2) Sound Interference
3) Beats
Announcements:
- Due to travel, no office hours tomorrow, Thursday
Last time: Sound Waves (longitudinal waves)
- Sound waves may be described as the displacement of individual particle - s(x,t)
- Sound waves may also be described in terms of variations of pressure at various points.
- Regions of high and low pressure correspond to places where the average displacement
of air is zero. Hence molecule displacement and pressure variation are π/2 rad (or 90o) out
of phase.
∆pm...pressure amplitude which is related to the displacement amplitude, sm , as follows:
∆pm = (vρω)sm
Speed of sound waves:
B
vsound , fluid 
. (in fluid or gas)

B ... Bulk modulus (measures how easy it is to compress the molecules in a medium)
Y
vsound , solid 
. (in solid)

Y... Young's modulus (is a measure of the stiffness of an elastic material)
So the sped of sound is generally larger in solids (B increases more than ρ)
Intensity of sound:
1
2
 v 2 sm2 [W/m ]
2
P
I s2
4 r
I
 I 
The Decibel Scale   (10dB )log  
 I0 
dB is the abbreviation for decibel, I0 is a standard reference intensity (= 10-12 W/m2),
chosen because it is near the lower
limit of the human range of hearing.
I is the intensity of the sound wave
PES 2130 Fall 2014, Spendier
Lecture 13/Page 2
1) Standing sound waves
Last lecture demo: Rubens' Flame Tube
- standing sound waves in tube
- wave pattern changed as we changes frequency of the sound
- we saw standing pressure waves only at specific resonant frequencies
Recall from Chapter 16 that standing waves can be set up on a stretched string that is
fixed at both ends.
- Waves traveling along the string are reflected back onto the string at each end.
- the superposition of waves traveling in opposite directions produces a standing wave
pattern (or oscillation mode).
- The wavelength required a standing wave
, where n = 1,2,3,….
is one that corresponds to a resonant frequency of the string.
fn  n
v
 nf1
2L
- the string then oscillates with a large, sustained amplitude, pushing back and forth
against the surrounding air and thus generating a noticeable sound wave with the same
frequency as the oscillations of the string (guitar)
We can set up standing waves of sound in an air-filled pipe in a similar way.
As sound waves travel through the air in the pipe, they are reflected at each end and
travel back through the pipe.
If the wavelength of the sound waves is suitably matched to the length of the pipe, the
superposition of waves traveling in opposite directions through the pipe sets up a
standing wave pattern. The wavelength required of the sound waves for such a match is
one that corresponds to a resonant frequency of the pipe.
Pipe with two open ends:
The reflection occurs even if pipe has open ends, but the reflection is not as complete as
when one end is closed.
At the ends of the pipe, the air pressure must equal the air pressure outside, ∆P=0.
PES 2130 Fall 2014, Spendier
Lecture 13/Page 3
Hence a plot of the change in pressure has nodes at the pipe ends and the air molecule
displacement plot is phase shifted by π/2 (where pressure fluctuations have a node,
average displacement must have an antinode)
We can compare this to the standing wave modes of a fixed string:
open pipe (two open ends):
, where n = 1,2,3,….
fn  n
v
 nf1
2L
Pipe with one end open:
At the closed end, the net displacement of air molecules must be zero. Therefore, the
displacement plot has a node at the closed end (like a fixed string) and antinode at the
open end (where molecules can move freely)
(Your turn: draw the corresponding pressure diagram)
PES 2130 Fall 2014, Spendier
Lecture 13/Page 4
As required, across the open end there is an antinode and across the closed end there is a
node. The simplest pattern requires sound waves having a wavelength given by L = λ/4,
so that λ = 4L. The next simplest pattern requires a wavelength given by L = 3 λ /4, so
that λ = 4L/3, and so on.
Note that only odd harmonics can exist in a pipe with one open end. For example, the
second harmonic, with n = 2, cannot be set up in such a pipe.
Pipe with one open end:
4L
4L
, where n' = 1,3,5, ...

(2n  1) n '
v
v
f n  (2n  1)
 n'
4L
4L
or
n 
f n  (2n  1) f1  n ' f1
Question:
A piccolo produces much higher frequencies than a saxophone. What does this tell you
about the relative sizes of the piccolo and saxophone?
Answer:
Both these instrument produce resonant frequencies based on the length of the pipe
(amongst other things) The resonant frequencies of the piccolo are higher and since
f  v / L , this means that the piccolo should be smaller than a saxophone
f  L  (inversely proportional)
PES 2130 Fall 2014, Spendier
Lecture 13/Page 5
2) Sound interference
Like transverse waves, sound waves can undergo interference. Sound waves coming from
multiple speakers will interfere. At some position away from the speaker you may hear a
loud sound and at other positions you may hear a soft or no sound. Why it this the case?
We know that when we add multiple waves (compute their superposition) we can have
construction, destructive, or intermediate interference.
Two speakers emit waves in phase:
At point P: The path length of the wave from both speakers is the same; sound from the
two speakers arrives in phase at point P.
At point Q: The path length of the wave from both speakers differs by λ/2; sound from
the two speakers arrive out of phase by 1/2 cycles at point Q.
The difference in path lengths means that the waves may not be in phase at any point.
In other words, their phase difference ϕ at any point depends on their
path length difference: L  L2  L1
recall that a phase difference of 2π rad corresponds to one wavelength
 L

2

L

 2 

Fully constructive interference occurs when   m(2 ) , for m = 0,1,2,....
L
 0,1,2,..... (fully constructive)

NOTE: a whole number of wavelength fits into the path difference)
PES 2130 Fall 2014, Spendier
Lecture 13/Page 6
Fully destructive interference occurs when   (2m  1)( ) , for m = 0,1,2,....
L
 0.5,1.5,2.5,..... (fully destructive)

NOTE: not a whole number of wavelength fits into the path difference)
Keeping this in mind is very important for a concert (multiple speakers) or indoor
surround sound system (reflection on smooth walls)
Example:
Two small loudspeakers, A and B, are driven by the same amplifier and emit pure
sinusoidal waves in phase. If the speed of sound is 350 m/s
(a) For what frequencies does constructive interference occur at point P?
(b) For what frequencies does destructive interference occur at point P?
PES 2130 Fall 2014, Spendier
Lecture 13/Page 7
3) Beats
Thus far we talked about interference that occurs when two different waves with the same
frequency overlap in the same region in space. Now, let's look what happens when we
have two waves with equal amplitude but slightly different frequencies. This occurs, for
example, when two organ pipes that are supposed to have exactly the same frequency are
slightly "out of tune." This situation creates so called "beats".
Beats occur when two waves with ALMOST the same frequency interfere. (Ignore spatial
variation.)
DEMO
Listen to two sounds whose frequencies are, 500 and 502 Hz, most of us cannot tell one
from the other. However, if the sounds reach our ears simultaneously, what we hear is a
sound whose frequency is 501 Hz, the average of the two combining frequencies. We
also hear a striking variation in the intensity of this sound—it increases and decreases in
slow, wavering beats that repeat at a frequency of 2 Hz, the difference between the two
combining frequencies. This is called the beat phenomenon.
Let's do a calculation: Let the time-dependent variations of the displacements due to two
sound waves of equal amplitude sm be
where ω1 > ω2 and ω1 ~ ω2
From the superposition principle, the resultant displacement is
Note:
1
 '  1  2 
2
1
  1  2 
2
is small!!!!
is approximately ω1 or ω2
PES 2130 Fall 2014, Spendier
Lecture 13/Page 8
So we have a wave with a small frequency (large amplitude) multiplied by a wave with
large frequency (small amplitude).
Since ω1 ~ ω2 we can then regard as a cosine function whose angular frequency is ω and
whose amplitude (which is not constant but varies with angular frequency ω') is the
absolute value of the quantity in the brackets.
Maximum amplitude:
cos(ω't) = +1 or -1
This happens twice in each repetition of the cosine function, so the beat occurs at
ωbeat = 2ω' = ω1 - ω2
since ω = 2πf
we get for the beat frequency:
fbeat = f1 - f2
Musicians use the beat phenomenon in tuning instruments.