Post on 27-Aug-2018
R. J. Wilkes
Email: ph116@u.washington.edu
Physics 116
Lecture 9 Standing waves
Oct 13, 2011 http://okamusic.com/
10/13/11 1 phys 116
•! 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.
Announcements
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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
Point A moves with big amplitude
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
Anti-node
Node
Nodes = stationary points; anti-nodes=maxima
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/s2 (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
–! 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
www.physicsclassroom.com/
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!
•! So resonant harmonics are L=!/4, 3!/4, 5!/4 … (1st, 3rd, 5th…)
–! 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
Closed Open
8
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
•! So resonant harmonics are L=!/2, 2!/2, 3!/2 … (1st, 2nd, 3rd…)
Open Open
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?
In general,
In this case,
•! Where should the fret be to make it resonate at E two octaves higher, at f =
1320 Hz?
L
L/2 L/4
Nut
Bridge
Frets
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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
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•! Interference pattern like this shows
locations of nodes and antinodes in space.
•! We can also create moving interference
patterns, so a stationary observer hears
cyclic intensity changes as maxima pass:
This is called beating, or a beat frequency
Beats Anti-node
Node
•! Beats are heard if waves with similar frequencies overlap at the observer’s
location, x: then at that spot, amplitude vs time is
The sum has a base frequency fOSC ,
modulated by an envelope of
frequency fBEAT
fOSC
fBEAT
Notice: fBEAT is twice the f in the cosine
function: Envelope goes from max to min
in ! cycle, so frequency of pulsation is
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