Physics 116 Lecture 9 Standing waves Oct 13, 2011 - - PowerPoint PPT Presentation

• 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

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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

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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

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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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