R. J. Wilkes

Email: ph116@u.washington.edu

Physics 116

Lecture 9 Standing waves

Oct 13, 2011 http://okamusic.com/

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•! 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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3

Today

Lecture Schedule (up to exam 1)

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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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Beats

•! Example: pluck 2 strings on a guitar: middle C (261 Hz) and G (392 Hz)

base frequency modulated by an envelope:

fOSC=327Hz

fBEAT=131 Hz

fOSC=327Hz = ~ E in next octave

fBEAT=131 Hz = C an octave below middle C

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