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Sound

Sound. Sound waves. Sound waves are longitudinal waves They cause particles of air to vibrate back and forth. A medium must be present in order for a sound wave to be heard. Properties of Sound Waves. Sounds waves have all the properties of any type of wave velocity Reflection Diffraction

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Sound

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  1. Sound

  2. Sound waves • Sound waves are longitudinal waves • They cause particles of air to vibrate back and forth. • A medium must be present in order for a sound wave to be heard.

  3. Properties of Sound Waves • Sounds waves have all the properties of any type of wave • velocity • Reflection • Diffraction • Interference • frequency • wave length • amplitude

  4. Properties of Sound Waves Velocity – 343m/s ( at room temp, 20 ° C) in air, increases with temperature of air. Temperature correction Formula- v = 331 m/s + 0.6(Temp) The velocity increases by 0.6 m/s for every degree the air temp. goes up. V of sound at 0 ° C = 331 m/s In gases, increases with air temperature. In solids and liquids, increases with density.

  5. Properties of Sound Waves Reflection – sound wave “bounces” off of a hard surface. - echoes - SONAR Example: 10 seconds after a gun is fired, the echo is heard. How far away is the wall of the canyon? (air temp = 20 ° C) ( it takes 5 sec. to reach the wall, and 5 sec. to return) v =d/t 334 m/s = d / 5 sec. d = 1670 m

  6. Properties of Sound Waves Diffraction - sound waves spread out after passing through a narrow opening. You can hear sounds from the hallway in my room, no matter where you sit.

  7. Properties of Sound Waves Interference – sound waves can build or cancel each other. - it can create “ dead spots” - can create “beats” that occur when two instruments are not in tune with each other. example – a 256 Hz tuning fork is struck along with a second tuning fork. A beat frequency of 3 Hz is heard. What are the 2 possible frequencies of the other tuning fork? answer - 253 Hz or 259 Hz

  8. Properties of Sound Waves Frequency - the pitch of a sound - still measured in Hertz - A high pitch has a high frequency and a short λ - A low pitch has a low frequency and a long λ - doppler shift An apparent shift in frequency due to motion

  9. Doppler Shift The 2 people hear 2 different frequencies. The driver hears the actual frequency.

  10. Doppler Effect Formula: f ’ = f(v + vd) v = speed of sound (v – vs) vd = speed of detector vs = speed of source f = actual frequency f’ = apparent frequency A car blowing it’s horn (f = 512 Hz) is approaching a stationary detector at a rate of 20 m/s. What frequency does the detector hear? assume air temp is 20 ° C f’= 512 Hz( 343 m/s + 0m/s) (343 m/s – (+20 m/s)) f ’ = 543 Hz

  11. Doppler Effect Formula: f’ = f(v + vd) v = speed of sound (v – vs) vd = speed of detector vs = speed of source f = actual frequency f’ = apparent frequency If source is approaching detector + vs If source is moving away - vs If detector is moving toward source +vd If detector is moving away - vd Frequency is always higher in front of moving object and lower behind it.

  12. Music • Resonance – air in an instrument is at the same frequency as the vibrating object. • It increases it’s amplitude due to constructive interference. • Pitch can be changed by changing the length of the air column • Open Pipe resonators • Closed Pipe resonators

  13. Open resonator Closed resonator

  14. Open Pipe Resonators A N A L = ½ λ or λ = 2L F1 F2 F3 A A L = 2/2L or L λ = 2/2/L or L L 3/2 L or λ = 2/3 L A A

  15. Closed Pipe Resonators A N L = 1/4 λ λ = 4L F1 F3 F5 A N A N L = 3/4 λ λ = 4/3L L = 5/4 λ λ = 4/5L A N

  16. Summary of Open and Closed Pipe Resonators Open Closed frequency fn = nf1 fn = nf1 n = 1,2,3,… n = 1,3,5,… Length L = n/2 λ L = n/4 λ Wavelength λ = 2/n L λ = 4/nL if speed of sound is known, f1 = v/2L` f1 = v/4L

  17. How to find the speed of sound lab As the water is lowered, resonance will occur each time the length of air is in resonance with the frequency of the tuning fork.

  18. Speed of Sound Lab A 493.9 Hz tuning fork is struck as a column of water is lowered. The first resonance is heard 0.16 m from the top of the tube. As the water is lowered, a second resonance is heard 0.505 m from the top. A third resonance is heard 0.860 m from the top. spacing 0.16 m 0.505 m 0.860 m λ = 2x spacing = 2(0.35 m) λ = 0.70 m v = f λ = (493.9 Hz)(0.70 m) = 346 m/s 0.345 m 0.345 m

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