1. 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

2. For a circular opening of diameter D
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)
For a circular opening of diameter D
These relations are useful in designing sound speakers D D

3. Example: Problem 17.14
A 3.00-kHz tone is being produced by a speaker with a diameter of 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 = m, T1 = 0 °C, T2 = 29 °C
Find: 1 and 2, to get = 2 -1
For circular speaker. Need s, therefore need v

4. For an ideal gas
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.

6. 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

7. Example: Problem 17.20 The beat frequency is In terms of periods
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?

8.  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.
If we chose fs smaller 

9. 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.

10. 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
If this time is equal to the period of the wave, T1, then the wave is a standing wave
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

11. Or three half-cycles Or n half-cycles Some notation:
The zero amplitude points are called nodes; the maximum amplitude points are the antinodes
For a string fixed at both ends