Diffraction
 What happens when a wave encounters an
obstacle or opening?
 It ``bends’’ around it.
 Consider a wave front (a wave in 2D) viewed
from above
Before the obstacle, the
wave front is planar (plane
wave). After the obstacle,
because of diffraction, the
wave fronts are circular
(spherical wave)
Obstacle
 However, the sound is mostly confined to within
an angle  to either side of the opening.  defines
the location of the first minimum in the sound
intensity
 For a rectangular slit opening of width D
(assuming the height to be very large)
sin  

D
D
 For a circular opening of diameter D
sin   1.22

D
D
 These relations are useful in designing sound
speakers
Example: Problem 17.14
A 3.00-kHz tone is being produced by a speaker
with a diameter of 0.175 m. The air temperature
changes from 0 to 29 °C. Assuming air to be an
ideal gas, find the change in the diffraction angle .
Solution:
Given: f = 3000 Hz, D = 0.175 m, T1 = 0 °C, T2 =
29 °C
Find: 1 and 2, to get = 2 -1
sin 
For circular speaker.
Need s, therefore need v
 1.22

D
v

f
For an ideal gas
v
 kT
m
Also, from Table 16.1, we know that the speed of
sound in air is 331 m/s (=v1) and 343 m/s for 0
°C and 20 °C, respectively. Therefore, we don’t
need to know , k, or m.
v1   kT1 / m , v2   kT2 / m
v2
T2
T2

 v 2  v1
v1
T1
T1

T1  T1 ( C )  273.15  0  273.15  273.15 K

T2  29 C  273.15  302.15 K
v 2  v1
T2
302.15 K
m
 (331 s )
 348 ms
T1
273.15K
sin  1  1.22
1
and 1  v1 / f
D
1 
v1 
1 
1 
  1  sin 1.22   sin 1.22 
D
fD 


and same for 2
v2 
v1 
1 
1 
   2   1  sin 1.22   sin 1.22 
fD 
fD 


348
331


1 
1 
 sin 1.22
  sin 1.22

3000* 0.175
3000* 0.175





 53.97 C  50.3 C  3.7 C
Beats
 Different waves usually don’t have the same
frequency. The frequencies may be much different or
only slightly different.
 If the frequencies are only slightly different, an
interesting effect results  the beat frequency.
 Useful for tuning musical instruments.
 If a guitar and piano, both play the same note
(same frequency, f1=f2)  constructive interference
 If f1 and f2 are only slightly different, constructive
and destructive interference occurs
 The beat frequency is
f b  f 2  f1 ( f 2  f1 ) or
1
1 1


In terms of periods
Tb T2 T1
as f 2  f 1 , f b  0
 The frequencies become ``tuned’’
Example: Problem 17.20
When a guitar string is sounded along with a 440Hz tuning fork, a beat frequency of 5 Hz is heard.
When the same string is sounded along with a 436Hz tuning fork, the beat frequency is 9 Hz. What is
the frequency of the string?
Solution:
Given: fT1=440 Hz, fT2=436 Hz, fb1=5 Hz, fb2=9 Hz
But we don’t know if frequency of the string, fs, is
greater than fT1 and/or fT2. Assume it is.
f b1  f s  f T 1 and f b 2  f s  f T 2 
f s  f b1  f T 1  5  440  445Hz
f s  f b 2  f T 2  9  436  445Hz
If we chose fs smaller
f b1  f T 1  f s and f b 2  f T 2  f s 
f s  f T 1  f b1  440  5  435Hz

f s  f T 2  f b 2  436  9  427 Hz
Standing Waves
 A standing wave is an interference effect due to
two overlapping waves
- transverse – wave on guitar string, violin, …
- longitudinal – sound wave in a flute, pipe
organ, other wind instruments,…
 The length (dictated by some physical constraint)
of the wave is some multiple of the wavelength
 You saw this in lab two weeks ago
 Consider a transverse wave (f1, T1) on a string of
length L fixed at both ends.
 If the speed of the wave is v (not the speed of
sound in air), the time for the wave to travel from
one end to the other and back is
2L / v
 If this time is equal to the period of the wave,
T1, then the wave is a standing wave
1 2L
v
v
T1 

 f1 

 1  2 L
f1
v
2 L 1
 Therefore the length of the wave is half of a
wavelength or a half-cycle is contained between
the end points
 We can also have a full cycle contained between
v
v
end points
2  L  f 2 
2

L
 f2
 Or three half-cycles
v
3v
3  L  f 3 
 2 
 f3
3 3 L 2 L
2
3
 Or n half-cycles
 Some notation:
v
 v 
f n  n , n  1, 2, 3, 4, ...
 2 L  For a string fixed
at both ends
f1
1st harmonic or fundamental
f 2  2 f1 2nd
1st overtone
f 3  3 f1 3rd
2nd overtone
f 4  4 f1 4th
3rd overtone
 The zero amplitude points are called nodes;
the maximum amplitude points are the antinodes