SOUND

1. SOUND A vibrating object, such as your voice box, stereo speakers, guitar strings, etc., creates longitudinal waves in the medium around it. When these waves cause our ear drums to vibrate, we “hear” sounds. Sound is caused byvibrations!

2. Sonic Spectrum • The frequency range over which longitudinal waves occur. • That part of the sonic spectrum that the human ear is sensitive to is called the aural range (a.k.a. sounds) • 20 Hz – 20,000 Hz (wavelength range: 17mm – 17m) • Ultrasonic: frequencies above 20,000 Hz • Infrasonic: frequencies below 20 Hz

3. Sonic Spectrum • Upper frequency limit is determined by the medium. • If the wavelength of the sound is small compared to the inter-particle spacing the wave will not be transmitted. • Small wavelength means high frequency. • Gases – around 109 Hz at ordinary temperature and pressure. • Solids/liquids – higher than 109 Hz due to closer spacing of particles.

4. Sound Waves • Wave energy is passed through the particles of the medium as a periodic, longitudinal wave. • Remember: the wave travels, not the medium! • The particles alternately experience compression and rarefaction.

5. Speed of Sound • Factors affecting the speed of sound • Temperature: the hotter the medium the faster the speed of sound. • Air: vsound = 331.5 m/s + .6T (T is in degrees C) • Density: the denser the medium the faster the speed of sound. • Air @ 00C: 331.5 m/s • Steel: 5200 m/s

6. Try one! • A ship sounds its fog horn to find out how far away an iceberg is. If the captain hears the echo 6 sec after sounding the horn, how many meters away is the iceberg? (assume T = 60 F) • Given: t = 6.0 s, T = 60 F 60F to Celsius: C = (5/9)(F – 32) = -14.40 C • Find the speed of sound: vsound = 322.9 m/s • Calculate the distance. Remember, the sound has to travel twice the distance between them! • 2d = vt so d = ((322.9m/s)*(6s))/2 = 969m

7. Sound Barrier • Sound waves spread out from their source in spherical shells, similar to ripples in a pond. • If the source is moving close to the speed of sound, the waves begin to pile up in front of it.

8. Sound Barrier • When the source moves at the speed of sound each new crest is created on top of the last one causing constructive interference • This creates an area of high pressure in front of the source called the “sound barrier.”

9. Super Sonic • Finally, when moving faster than the speed of sound, the source outruns the wave crests it creates. • The V pattern created by the successive wave crests is called a “shock wave,” with the source ahead of it. • This shock wave is the “sonic boom” we hear when something goes by at supersonic speed.

10. Properties of Sound • Pitch (frequency) • High pitch (high frequency): shorter wavelength sounds such as a siren or a flute. • Low pitch (low frequency): longer wavelength sounds such as a sub-woofer or a fog horn. • Pitch can also be described with musical notes. • Pitch (frequency) does not change when a sound wave passes from one medium to another. • Normal speech range; 1000 – 5000 Hz

11. Properties of Sound • Intensity (loudness) • A measure of the amount of sound energy that passes through a given area over a given time • I = P/A = Watts/cm2 • Power is determined by the source, but the area increases with the square of the distance from the source. • Intensity is inversely proportional to the square of the distance from the source • A sound heard from 100m away is ¼ as intense when heard from 200m away. • Intensity is also a measure of the amplitude of the sound wave.

12. Intensity (continued) • The decibel scale relates sound intensity to our perception of how loud sounds are. • Units: decibels (dB) • β = 10log(I/Io) • I is the intensity of the sound heard • Io is the intensity of a sound at the threshold of hearing Io = 1.00x10-12 W/m2 • β = 0 dB is the threshold of hearing. • β = 110 dB is considered the threshold of pain. • However, sounds below 110 dB can still cause hearing loss.

14. Try it! • The intensity of a sound is found to be 1x10-14 W/cm2. What is the sound level? • Given: I = 1x10-14 W/cm2 • Convert Io to W/cm2, Io = 1x10-16 W/cm2 • β = 10log[(1x10-14 W/cm2)/(1x10-16 W/cm2)] = 20 dB • How much would the sound level change if the intensity was doubled? • β would increase 3 dB.

15. Doppler Effect • The change in frequency of a sound due to the relative motion between the source and listener

16. A decreasing distance between the source and observer will cause a higher pitch to be heard. fo = frequency heard by observer fs= frequency of the source v = speed of sound vo = speed of the observer vs = speed of the source Doppler Effect

17. An increasing distance between the source and observer will cause a lower pitch to be heard. fo = frequency heard by observer fs= frequency of the source v = speed of sound vo = speed of the observer vs = speed of the source Doppler Effect

18. Try it! • A driver travels northbound on a highway at a speed of 25.0 m/s. A police car, traveling southbound at a speed of 40.0 m/s, approaches with its siren sounding at a frequency of 2,500 Hz. What frequency does the driver hear as the police car approaches? • Given: fs = 2,500 Hz, v = 343 m/s, vs = 40 m/s, vl = 25 m/s • The cars are getting closer so

19. Doppler Effect (Light) • A similar effect, but the equation is slightly different • fo = frequency seen by observer • fs = frequency of source • c = speed of light • v = relative speed between observer and source

20. More Properties of Sound • Reflection • Refraction • Interference • Need 2 waves • same frequency • in phase • When the waves travel to the same point, the difference in their path lengths determines what type of interference occurs

21. L1 L2 1D Interference • If L2 – L1 = n(½) (for n = 1,2,3…) then there will be total destructive interference • If L2 – L1 = m (for m =1,2,3…) then there will be total constructive interference

22. 2D Interference • As with 1D • L = n(½) results in total destructive interference • L = m results in total constructive interference • You have to rely more on the path lengths than visual cues for 2D

23. Beats • When two tones are heard at the same time they interfere with each other causing a pulsing sound called beats. • The frequency of the beat pattern is the difference between the frequencies of the two tones. • fb = f1 - f2

24. Let’s try it! • Jane holds two slightly different tuning forks next to her ear. What is the beat frequency she hears if one tuning fork vibrates at 440Hz and the other at 436Hz? fb = 440Hz – 436Hz = 4Hz

25. Shake it Up • Forced Vibration • a vibrating object is touched to a second object • the second object begins to vibrate at the same frequency • Resonance • a vibration caused in a medium due to a disturbance that occurs at the medium’s natural frequency • the natural frequency is the frequency of oscillation in an object that will produce a standing wave • unrestricted, the amplitude of the vibrations will continue to increase

26. The Sound of Music • Musical instruments are designed to resonate at one or more natural frequencies • The strings on a string instrument have a base frequency based on its length and tension (remember last chapter?), but they can produce other notes by putting pressure on the fret board effectively shortening the string • Wind and brass instruments create resonance patterns in the “pipe-like” body

27. Resonance in Pipes • There are two kinds of pipes to consider • Open pipes – open at both ends • a standing wave is created in the pipe such that there is an antinode at each end • the wavelength of the standing wave depends on the length of the pipe for n = 1,2,3,…

28. Resonance in Pipes • Closed pipes – closed at one end open at the other • a standing wave is created in the pipe such that there is a node at the closed end and an antinode at the open end • the wavelength of the standing wave depends on the length of the pipe for n = 1,3,5

29. Wind and Brass Instruments • Wind instruments act like an open pipe • Brass instruments act like closed pipes

30. Let’s try it! • In an unheated 100C room, a 25cm pipe is producing its 3rd harmonic. If the pipe is open at both ends, what is the frequency of the tone heard? first you find the speed of sound in the room v = 331.5 + .6(10) = 337.5m/s then you use the open pipe equation for n = 3 f = 3(337.5/(2x0.25m)) = 2025Hz