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 the location of the first minimum in the sound intensity  to either side of the opening.  defines  For a rectangular slit opening of width D (assuming the height to be very large) sin   

D

For a circular opening of diameter D 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, T 1 29 °C = 0 °C, T 2 = Find:  1 and  2, to get  =  2  1 For circular speaker.

Need  s, therefore need v sin   1 .

22 

D

 

f

v 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 (=v °C and 20 °C, respectively. Therefore, we don’t need to know  , k, or m .

1 ) and 343 m/s for 0 v 1  

kT

1 /

m

, v 2  

kT

2 /

m

v v 1

T

1

T

2 2 

T T

1 2  v 2  v 1

T

2  

T

1 ( 29  

C C

)   273 .

15 273 .

15  

T

1 0  273 302 .

15 K .

15  273 .

15 K

v  2 sin    1 1 v   1

T

2 1 .

22 sin

T

1  1  ( 331 m   1 and

D

   1 .

22 

D

1 s )  1     302 .

15 K 273.15

K  sin v 1  1 /   

f

1 .

22  348 v 1

fD

   m s and     same sin 53  .

  1 97    2 1  .

for   22 C  1  2 3000 50 .

sin 3  *  1 348 0 C    .

1 .

22 175  3 .

7    v

fD

 2    sin   1 sin  1       1 .

22 1 .

22 3000 v 1

fD

331 * 0    .

175  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 interesting effect results  slightly the different, an beat frequency.

 Useful for tuning musical instruments.

 If a guitar and piano, both play the same note (same frequency, f 1 =f 2 )  constructive interference  If f 1 and f 2 are only slightly different, constructive and destructive interference occurs

 The beat frequency is

f b



f

2 

f

1 (

f

2 

f

1 ) or  1

T b

as 

f

1

T

2  

f

1

T

1 ,

f

In terms of periods  0 2 1

b

The frequencies become ``tuned’’ Example: Problem 17.20

When a guitar string is sounded along with a 440 Hz tuning fork, a beat frequency of 5 Hz is heard. When the same string is sounded along with a 436 Hz tuning fork, the beat frequency is 9 Hz. What is the frequency of the string?

Solution: Given: f T1 =440 Hz, f T2 =436 Hz, f b1 =5 Hz, f b2 =9 Hz But we don’t know if frequency of the string, greater than f T1 and/or f T2 . Assume it is.

f f f f f f b

1

s s b

2 If we chose f s

b

1

s s

     

f f f f f T T f b

1

T s

1 2 1      

f f f f T

1 and   5 9

T

2 smaller

f T

1 

f b

2   440 436

b

1

b s

2 and  

f

440

b

436  2   5 9

f

 

s

 445

f

445

T

2 Hz f s , is  Hz

f

 

T

2  435 Hz 427

f s

 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 length L transverse wave ( fixed at both ends. f 1 , T 1 ) on a string of

 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 2

L

/ v  If this time is equal to the period of the wave, T 1 , then the wave is a standing wave

T

1  1  2

L



f

1  v   v   1  2

L f

v 2

L

1 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 end points  2 

L



f

2   v 2  v

L



f

2

  Or three half-cycles 3  2 3

L



f

3   v 3  Or n half-cycles

f n

 

n

 2 3 v

L

v 2

L

  Some notation: 3v 2

L



f

3 ,

n

 1 , 2, 3, 4, For a string fixed at both ends ...

f f f f

1 3 4 2  2

f

1   3 4

f f

1 1 1st harmonic 2nd 3rd 4th or fundamenta 1st 2nd 3rd overtone overtone overtone l  The zero amplitude points are called the maximum amplitude points are the nodes ; antinodes