Physics 116 Lecture 9 Standing waves Oct 13, 2011 http://okamusic.com/ R. J. Wilkes Email: ph116@u.washington.edu 10/13/11 phys 116 1
Announcements • � HW2 (ch. 14) due today 5 pm • � HW3 open at 5pm (Ch.25, due 10/24) -- But focus on studying chs. 13-14 until Monday! • � Exam 1 is Monday, 10/17 • � All multiple choice, similar to HW problems • � YOU must bring a standard mark-sense (bubble) sheet • � Closed book/notes, formula page provided • � You provide: bubble sheet, pencils, calculator, brain • � Kyle Armour will hold a special office hour 11:30-12:30 on Monday 10/17 in room B442. • � Covers material in Chs.13 and 14 (through today’s class only) • � Damped/driven oscillators will NOT be on test • � Tomorrow’s class: review • � Example questions similar to test items AND formula page posted : see class home page. • � Try them before class tomorrow, when we will go over solutions. 10/13/11 phys 116 2
Lecture Schedule (up to exam 1) Today 10/13/11 phys 116 3
Self-interference: “standing waves” • � If I wiggle the rope at just the right f – � Waves reflected from the end interfere constructively with new waves I am making – � Result: looks as if some points stand still: standing waves • � Example of resonance: rope length L = multiple of ! /2 Anti-node Node Point A moves with big amplitude Nodes = stationary points; anti-nodes=maxima Point B has amplitude ~0 • � Same thing happens in musical instruments – � Structure favors waves which have L = multiple of ! /2 • � Guitar, violin strings: both ends must be nodes – � Organ pipes, wind instruments: one end must be node, other antinode
Vibrating strings • � Example: First string of guitar (thinnest) is tuned to E above middle C, f = 330 Hz • � String has length L=0.65m and mass 2 grams. What tension is needed? • � Frequency f=v/ ! � – � to get f=330 Hz for ! =L=0.65m, we need v=f ! =(330Hz)(0.65m) = 214 m/s • � Mass density of string is " = 0.002kg / 0.7m = 0.0028 kg/m • � v 2 = F/ " , so F= v 2 " = (214m/s) 2 0.0028kg/m = 128 kg-m/s 2 (notice: units=newtons) 4.5N=1lb, so about 28 lbs tension needed
Guitar strings • � Excite waves by plucking guitar string – � Waves with ! such that L = multiple of ! /2 are reinforced (resonate) • � L = ! /2, 2( ! /2), 3( ! /2)… all work: harmonics www.physicsclassroom.com/ – � What determines ! ? Speed of wave on string depends on • � Tension in string (force stretching it) – � Taut = higher speed, slack = lower speed • � Inversely on mass per unit length of string material – � Heavy string = slower speed, light string = faster So v = � F/ " , F is in newtons and " = kg/meter
Organ pipes • � Organ pipes have one closed and one open end • � Closed end must be a node, open end must be anti-node – � So L must be multiple of half of ! /2: L=N( ! /4) • � But if N=even number, we’d get two nodes: so N = odd # only! Closed Open • � So resonant harmonics are L= ! /4, 3 ! /4, 5 ! /4 … (1 st , 3 rd , 5 th …) – � Imitate organ-pipe operation by blowing across end of a bottle • � Put water in bottle to change fundamental frequency • � Brass and woodwind instruments work like organ pipes – � Use valves to change effective length (brass), or impose an antinode somewhere inside (woodwinds) • � Your ear canal is an example of a closed-end pipe
Open ended pipes • � Some instruments have both ends open: folk flutes (panpipes, Asian flutes), didgeridoo, etc • � Now both ends must be anti-nodes – � L should be integer multiple of ! /2: L=n ( ! /2) • � But now n=even number works also – � so frequencies are same as for guitar strings Open Open • � So resonant harmonics are L= ! /2, 2 ! /2, 3 ! /2 … (1 st , 2 nd , 3 rd …) 8
Examples • � Guitar string is tuned to E, at f = 330 Hz when open (L= nut to bridge) • � Where should a fret be placed to make it resonate at E one octave higher, at f = 660 Hz? Bridge In general, Frets Nut L L/2 In this case, L/4 • � Where should the fret be to make it resonate at E two octaves higher, at f = 1320 Hz? 10/13/11 phys 116 9
Examples • � According to fig 14-28, the first harmonic of the human ear canal is located at f=3500 Hz • � If we model the ear canal as a simple organ pipe, how long must it be? f=3500 Hz 10/13/11 phys 116 10
Beats • � Interference pattern like this shows Anti-node locations of nodes and antinodes in space. • � We can also create moving interference Node patterns, so a stationary observer hears cyclic intensity changes as maxima pass: This is called beating, or a beat frequency • � Beats are heard if waves with similar frequencies overlap at the observer’s location, x: then at that spot, amplitude vs time is f OSC f BEAT The sum has a base frequency f OSC , modulated by an envelope of Notice: f BEAT is twice the f in the cosine frequency f BEAT function: Envelope goes from max to min in � cycle, so frequency of pulsation is 10/13/11 phys 116 11
Beats • � Example: pluck 2 strings on a guitar: middle C (261 Hz) and G (392 Hz) f OSC =327Hz f BEAT =131 Hz base frequency modulated by an envelope: f OSC =327Hz = ~ E in next octave f BEAT =131 Hz = C an octave below middle C 10/13/11 phys 116 12
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