PHYS2 Physics 1 FUNDAMENTALS Module 3 OSCILLATIONS & WAVES Text Physics by Hecht Chapter 11 WAVES Standing Waes Doppler Eect Sections: 11. 11.11 Examples: 11.12 11.13 11.14 11.15 CHECKLIST Standing Waes (stationary waes) intererence, nodes, antinodes, waelength λ is twice the node-to-node distance or twice the antinode-to-antinode distance Standing Waes on Strings - string ixed at both end undamental, harmonics, oertones, modes o ibration (Fig. 11.45) Node Antinode N A N A N λ FT 1 FT L = N N = N 1 = λ = 1 = µ = m/ L 2 u 2L u mode number, N = 1, 2, 3, (Eq. 11.14, 11.15) Standing Waes in air columns - both ends open or closed undamental, harmonics, oertones, modes o modal patterns, chamber open at both ends and closed at both ends pressure: open N A N A N open closed A N A N A closed particle displacement: open A N A N A open λ L = N N = N 1 = λ 2 mode number, N = 1, 2, 3, closed N A N A N closed (Eq. 11.14) a3\p1\waes\waes1111.doc 9:14 AM 1
Standing Waes in air columns - one end open, other closed undamental, harmonics, oertones, modes o ibration, modal patterns, chamber open / closed pressure: open N A N A closed particle displacement: open A N A N A closed λ L = N N = N 1 = λ (Eq. 11.16) 4 mode number, N = 1, 3, 5, only odd harmonics resonant Doppler Eect o = s ± o ± s (Eq. 11.22) o > s source and/or obsering moing towards each other only o < s source and/or obserer receding rom each other only Shock waes NOTES STANDING (STATIONARY) WAVES Wae traelling to right ---> y 1 (x,t) = A sin(k x - ω t) Wae traelling to the let <--- y 2 (x,t) = A sin(k x + ω t) Two waes traelling in opposite directions with equal displacement amplitudes and with identical periods and waelengths interere with each other to gie a standing (stationary) wae (not a traelling wae - positions o nodes and antinodes are ixed with time) standing wae (no proo) y R (x,t) = 2 A sin(k x) cos(ω t) amplitude o standing wae (aries with position only and is independent o time) y Rm (x) = 2 A sin(k x) cos(ω t) each point oscillates with SHM, period T = 2π / ω a3\p1\waes\waes1111.doc 9:14 AM 2
Nodes I k x = n π (n =, 1, 2, 3,...) then y R (x) = zero amplitude at x and at all times t k = 2π / λ (2π x) / λ = n π x = n λ / 2 adjacent nodes are separated by λ / 2 Antinodes I k x = (n + ½) π (n =, 1, 2, 3,...) then y R (x) = 2A max amplitude at x and at all times t k = 2π / λ (2π x) / λ = (n + ½) π x = (n + ½)λ / 2 adjacent antinodes are separated by λ / 2 and are located halway between pairs o nodes STANDING WAVES IN STRINGS AND RESONANCE String instruments orm a large group o musical instruments which include the iolin, guitar and piano. All these instruments make a sound by causing a taut string to ibrate. The string may be bowed (iolin), plucked (guitar) or struck by a hammer (piano). The pitch o the note produced depends on three actors length, linear density and string tension. A shorter, lighter or tighter string gies a higher note. The iolin amily o instruments are the most expressie o sting instruments. The iolin has our strings o dierent linear densities. These are wound around tuning pegs to produce the correct tension. The perormer stops the string ibrating to obtain other notes. By pressing one or more strings against the ingerboard to shorten the section that ibrates, higher notes can be played. The body o the iolin acts as an resonant ampliier. The ront and back o the iolin are connected by a short sound post that transmits ibrations to the back. The whole body ibrates and the sound wae is emitted through -shaped sound holes on the ront o the instrument. a3\p1\waes\waes1111.doc 9:14 AM 3
Waes relected at boundaries O------------------------------------O x = x = L Boundary conditions imposed on ibrating string y R (,t) = and y R (L,t) = The natural ibrations o the string are described by y R (x,t) = 2 A sin(k x) cos(ωt) At x = y R (,t) = boundary condition satisied At x = L y R (L,t) = boundary condition k L = N π N = 1, 2, 3,... N is reerred to as the mode number The waelength λ is determined by the distance L between the supports at the end 2L λn o the string λn = L= N N 2 The speed o a transerse wae on a string is determine by the string tension and its linear density = Natural requencies FT µ N 1 FT = = N λ 2L µ N Resonance ( large amplitude oscillations) occurs when the string is excited or drien at one o its natural requencies. Harmonic series N = 1 undamental or irst harmonic λ 1 = 2L 1 = (1/2L). (F T / µ) N = 2 2nd harmonic (1st oertone) λ 2 = L = λ 1 / 2 2 = 2 1 N = 3 3rd harmonic (2nd oertone) λ 3 = 2L / 3 = λ 1 / 3 3 = 3 1 N Nth harmonic (N-1 th oertone) λ N = 2L / N = λ 1 / N N = N 1 N 12 8 6 4 3 4 2 2.1.2.3.4.5.6.7.8.9 1 position along string 1 a3\p1\waes\waes1111.doc 9:14 AM 4
Why does a iolin sound dierent to a iola? Why do musicians hae to tune their string instruments beore a concert? Fingerboard Body o instrument (belly) resonant chamber - ampliier dierent string - µ bridges - change L tuning knobs (pegs) - adjust F T λ L F = T 1 FT µ L µ 1 F 2L µ T 1 = N = N 1 N = 1,2,3,... iolin spectrum 9 5 4 8 3 7 2 6 5 4-3 -2 2 1 1 2 3 4 5 6 7 8 9 11 12 13 14 15 harmonics (undamental 1 = 44 Hz) -3-4 -5.2.4.6.8 time t (s) iola spectrum 14 5 12 4 3 2 8 6 - -2 4-3 2-4 -5 1 2 3 4 5 6 7 8 9 11 12 13 14 15 harmonics (undamental 1 = 44 Hz).2.4.6.8 time t (s) a3\p1\waes\waes1111.doc 9:14 AM 5
Problem A particular iolin string plays at a requency o 44 Hz. I the tension is increased by 8.%, what is the new requency? Solution A = 44 Hz B =? Hz F TB = 1.8 F TA λ A = λ B µ A = µ B L A = L B N A = N B = λ = (F T / µ) string ixed at both ends L = N λ/2 λ = 2L / N natural requencies N = N / 2L = (N / 2L). (F T / µ) B / A = (F TB / F TA ) 2B = (44) (1.8) = 457 Hz Problem A guitar string is 9 mm long and has a mass o 3.6 g. The distance rom the bridge to the support post is 6 mm and the string is under a tension o 52 N. 1 Sketch the shape o the wae or the undamental mode o ibration 2 Calculate the requency o the undamental. 3 Sketch the shape o the string or the sixth harmonic and calculate its requency. 4 Sketch the shape o the string or the third oertone and calculate its requency. Solution L 1 = 9 mm =.9 m L = 6 mm =.6 m m = 3.6 g = 3.6-3 kg F T = 52 N µ = m / L 1 = 3.6-3 /.9 =.4 kg.m -1 = (F T / µ) = (52 /.4) = 36.6 m.s -1 λ 1 = 2L = (2)(.6) = 1.2 m Fundamental requency 1 = / λ 1 = 36.6 / 1.2 = 3 Hz N = N 1 sixth harmonic N = 6 6 = (6)(3) = 18 Hz = 1.8 khz third oertone = 4th harmonic N = 4 4 = (4)(3) = 12 = 1.2 khz a3\p1\waes\waes1111.doc 9:14 AM 6
STANDING WAVES IN AIR COLUMNS Woodwind instruments are not necessarily made o wood eg saxophone, but they do require wind to make a sound. They basically consist o a tube with a series o holes. Air is blow into the top o the tube, either across a hole or past a lexible reed. This makes the air inside the tube ibrate and gie out a note. The pitch o the note depends upon the length o the tube. A shorter tube produces a higher note, and so holes are coered. Blowing harder makes a louder sound. To produce deep notes woodwind instruments hae to be quite long and thereore the tube is cured. Brass instruments (usually made o brass) consist o a long pipe that is usually coiled and has no holes. The player blows into a mouthpiece at one end o the pipe, the ibration o the lips setting the air column ibrating throughout the pipe. The trombone has a section o pipe called a slide that can be moed in and out. To produce a lower note the slide is moed out. The trumpet has three pistons that are pushed down to open extra sections o tubing. Up to six dierent notes are obtained by using combinations o the three pistons. For pipes closed at both ends or open at both ends: all harmonics exists just like a string ixed at both ends. Pipe closed at one end and open at the other closed end particle displacement zero node open end max particle displacement antinode node Particle displacement zero antinode Particle displacement maximum a3\p1\waes\waes1111.doc 9:14 AM 7
12 8 6 4 2.1.2.3.4.5.6.7.8.9 1 position along column Boundary conditions - relection o sound wae at end o air column (pipe) At an open end a compression is relected as a rareaction and a rareaction as a compression (π phase shit). To match the boundary conditions 4L λ2n 1 λ2n 1 = L= ( 2N 1) N = 1,2,3,... 2N 1 4 Speed o sound in air (at room temperature ~ 344 m.s -1 ) = λ Natural requencies o ibration (open closed air column) 2N 1 2N 1 = = λ 4L 2N 1 odd harmonics exit: 1, 3, 5, 7, a3\p1\waes\waes1111.doc 9:14 AM 8
14 13 12 11 9 8 7 6 5 4 3 2 1-1.1.2.3.4.5.6.7.8.9 1 equilibrium position o particles instantaneous position o particles sine cure showing instantaneous displacement o particles rom equilibrium instantaneous pressure distribution time aeraged pressure luctuations Enter t/t An air stream produced by mouth by blowing the instruments interacts with the air in the pipe to maintain a steady oscillation. All brass instruments are closed at one end by the mouth o the player. Flute and piccolo open at atmosphere and mouth piece (embouchure) coering holes L λ Trumpet open at atmosphere and closed at mouth coering holes adds loops o tubing into air stream L λ Woodwinds ibrating reed used to produce oscillation o the air molecules in the pipe. a3\p1\waes\waes1111.doc 9:14 AM 9
Problem What are the natural requencies o a human ear? Why do sounds ~ 3 4 Hz appear loudest? Solution Assume the ear acts as pipe open at the atmosphere and closed at the eardrum. The length o the auditory canal is about 25 mm. Take the speed o sound in air as 34 m.s -1. L = 25 mm =.25 m = 34 m.s -1 For air column closed at one end and open at the other L = λ 1 / 4 λ 1 = 4 L 1 = / λ 1 = (34)/{(4)(.25)} = 34 Hz When the ear is excited at a natural requency o ibration large amplitude oscillations (resonance) sounds will appear loudest ~ 3 4 Hz. a3\p1\waes\waes1111.doc 9:14 AM
DOPPLER EFFECT - motion related requency changes Doppler 1842, Buys Ballot 1845 - trumpeters on railway carriage o = s ± ± o s obserer (source) = (source) = (wae) (source) = (wae) / 2 (source) = 1.25 (wae) source s obserer o obsered requency o stationary stationary = o stationary receding < o stationary approaching > o receding stationary < o approaching stationary > o receding receding < o approaching approaching > o approaching receding? receding approaching? a3\p1\waes\waes1111.doc 9:14 AM 11
Applications: police microwae speed units, speed o a tennis ball, speed o blood lowing through an artery, heart beat o a deeloping etous, burglar alarms, sonar ships & submarines to detect submerged objects, detecting distance planets, obsering the motion o oscillating stars. Shock Waes supersonic waes boat moing through water: speed o boat > speed o water wae created ast moing power boat sailing boat rocket iolently - wae crests add to gie large wae shock wae - bunching o waeronts ---> abruptie rise and all o air pressure Mach cone sonic boooom plane traelling at supersonic speeds Mach Number = / s s speed o sound in air eg Mach Number = 2.3 speed is 2.3 times the speed o sound a3\p1\waes\waes1111.doc 9:14 AM 12
Problem The speed o blood in the aorta is normally about.3 m.s -1. What beat requency would you expect i 4. MHz ultrasound waes were directed along the blood low and relected rom the end o red blood cells? Assume that the sound waes trael through the blood with a elocity o 154 m.s -1. Solution Setup s1 = 4. MHz = 4.x 6 Hz 1 =.3 m.s -1 = 1.54x 3 m.s -1 o1 =? Hz o2 =? Hz s2 = o1 relected wae s2 =.3 m.s -1 Doppler Eect Action o = s ± ± o s Beats beat = 2 1 Blood is moing away rom source obserer moing away rom source o < s ( ) 4 3.999221 Hz 3 ± o1 6 1.54.3 6 o1 = s1 = 3 = ± s1 1.54 Wae relected o red blood cells source moing away rom obserer o < s ( ) 3.999221 3.998442 Hz 3 ± o2 6 1.54 6 o2 = s2 = 3 = ± s2 1.54 +.3 Beat requency = 4. 3.998442 6 Hz = 1558 Hz In this type o calculation you must keep extra signiicant igures. a3\p1\waes\waes1111.doc 9:14 AM 13