sound is a longitudinal, mechanical wave that travels as a series of high and low pressure variations

Transcription

1 Sound

2 sound is a longitudinal, mechanical wave that travels as a series of high and low pressure variations

3 the high pressure regions are compressions and the low pressure regions are rarefactions the speed of sound depends upon the temperature and the density of the medium

4 Sound can be pictured as a transverse wave as pressure vs time

5 Mythbusters

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7 Resonance in Air Columns Columns of air can resonate at more than 1 frequency Standing waves of sound can be set up in a column of air by sound waves traveling in opposite directions

8 When a sound source is held over the open end of a tube, the sound wave travels down the tube and reflects off the closed end If the reflected and the incident waves meet in phase, constructive interference will occur

9 If the reflected and the incident waves meet in phase, constructive interference will occur A node will form at the closed end (no displacement of air) and an antinode will form at the open end

10 Closed Tube The longest wavelength that will be resonant = ¼ the L length of the tube 4

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12 Lengthening Tube the tube will be resonant whenever a node occurs at the closed end and an antinode occurs at the open end

13 the tube will be resonant whenever a node occurs at the closed end and an antinode occurs at the open end

14 a pipe open at both ends will have an antinode at each end (example: flutes) Open tubes since antinodes are ½ apart, the fundamental resonant length is given by L 2

15 Example A column of air closed at one end is m long. What is the lowest resonant frequency of the pipe? (Speed of sound is 343 m/s)

16 L v=f f 4 =4L = 1.42 m (longest wavelength that can fit into the pipe) v f 343 ms / 1.42 m f =242 Hz

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20 Example A closed tube resonates at 440 Hz when it is m long and again at m long. Determine the speed of sound.

21 Solution Experiments have shown that the antinode is above the open end of the tube. The extra length depends on the diameter of the tube. This can be corrected for by subtracting the resonant lengths of the tube at a given frequency: ¾ - ¼ = ½.

22 Solution 4 3 L L L L m m m v f v = 344 m/s

23 Practice An open pipe 1.71 m long is resonating at a certain frequency. There are 3 antinodes in the pipe. If the speed of sound is 341 m/s, what is the resonant frequency?

24 Solution 3 antinodes in 1.71 m Must have antinodes at the open ends one in the centre The tube is operating at the second harmonic

25 Solution 3 antinodes in 1.71 m Must have antinodes at the open ends one in the centre L 1.71m 1.71m m f 341 m/ s 0.855m

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27 Solution 3 antinodes in 1.71 m Must have antinodes at the open ends one in the centre L 1.71m 1.71m m f 341 m/ s 0.855m

28 Diffraction & Sound diffraction is important in loudspeaker design. Speakers that produce low frequency bass notes are called woofers. They need to move a lot of air, so need to be quite large. Diameters of 30 cm or more are common.

29 a typical high note has a frequency of 5000 Hz or so, = 7cm. if you use the woofer to generate a high frequency note, the sound wave will beam straight ahead without significant diffraction, just like water waves passing through a wide gap. you won t hear those notes unless you are right in front of the speaker.

30 When the opening, max diffraction occurs (tweeter emitting high notes) When < opening size, little diffraction occurs (woofer emitting high notes)

31 Hearing the volume of the sound depends on the difference in the pressure between the compression and the rarefaction the human ear can detect sound from about 20 Hz to about 20 khz

32 The loudness of sound is determined by the intensity, the amount of energy that passes through a 1 m 2 area in 1 second. The volume (intensity) you hear depends on the square of the distance from the source Intensity 1 2 distance

33 The volume of sound is sometimes measured in decibels, db db compares the intensity of a sound to a reference sound Uses powers of 10

34 Source Intensity Intensity Level # of Times Greater Than TOH Threshold of Hearing (TOH) 1x10-12 W/m 2 0 db 10 0 Rustling Leaves 1x10-11 W/m 2 10 db 10 1 Whisper 1x10-10 W/m 2 20 db 10 2 Normal Conversation 1x10-6 W/m 2 60 db 10 6 Busy Street Traffic 1x10-5 W/m 2 70 db 10 7 Vacuum Cleaner 1x10-4 W/m 2 80 db 10 8 Large Orchestra 6.3x10-3 W/m 2 98 db MP3 at Maximum Level 1x10-2 W/m db Front Rows of Rock Concert 1x10-1 W/m db Threshold of Pain 1x10 1 W/m db Military Jet Takeoff 1x10 2 W/m db Instant Perforation of Eardrum 1x10 4 W/m db 10 16

35 Sound Energy How long would you have to yell in order to heat a cup of coffee?

36 Data (Assumptions?) Mass of coffee = 0.25 kg Energy/s of typical yell = 80 db J/s Change in temperature 50 o C Heat capacity of coffee 4200 J/kg o C Other Assumptions?

37 Solution Q = mct P E t Q = J to heat the coffee t E P t s t 601 days