HW12, Due Friday, May 8 th Thursday, Finish up, begin ... Due Friday, May 8 th Thursday, Finish up,...

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Physics 207 – Lecture 29

Physics 207: Lecture 29, Pg 1

Lecture 29Goals:Goals:

•• Chapter 20Chapter 20� Work with a few important characteristics of sound waves.

(e.g., Doppler effect)

•• Chapter 21Chapter 21� Recognize standing waves are the superposition of two

traveling waves of same frequency� Study the basic properties of standing waves� Model interference occurs in one and two dimensions� Understand beats as the superposition of two waves of

unequal frequency.

•• AssignmentAssignment� HW12, Due Friday, May 8th

� Thursday, Finish up, begin review for final, evaluations


Doppler effect, moving sources/receivers

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� If the source of sound is moving� Toward the observer ⇒

λ seems smaller

� Away from observer ⇒λ seems larger

� If the observer is moving� Toward the source ⇒

λ seems smaller

� Away from source ⇒λ seems larger

vvs

ff

−=

1source

observer

sourceo

observer vv

1 ff

+=

Doppler Example Audio

Doppler Example Visual

vvs

ff

+=

1source

observer

sourceo

observer vv

1 ff

−=


Doppler Example � A speaker sits on a small moving cart and emits a short 1

Watt sine wave pulse at 340 Hz (the speed of sound in air is 340 m/s, so λ = 1m ). The cart is 30 meters away from the wall and moving towards it at 20 m/s.

� The sound reflects perfectly from the wall. To an observer on the cart, what is the Doppler shifted frequency of the directly reflected sound?

� Considering only the position of the cart, what is the intensity of the reflected sound? (In principle on would have to look at the energy per unit time in the moving frame.)

t0

A

30 m

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Doppler Example� The sound reflects perfectly from the wall. To an observer on

the cart, what is the Doppler shifted frequency of the directly reflected sound?

At the wall: fwall = 340 / (1-20/340) = 361 Hz

Wall becomes “source” for the subsequent part

At the speaker f ’ = fwall (1+ 20/340) = 382 Hz

t030 m

t1

vvs

ff

−=

1source

observer

sourceo

observer vv

1 ff

+=


Example Interference

� Considering only the position of the cart, what is the intensity of the reflected sound to this observer? (In principle one would have to look at the energy per unit time in the moving frame.)

vcart ∆t + vsound ∆t = 2 x 30 m = 60 m

∆t = 60 / (340+20) = 0.17 s � dsound = 340 * 0.17 m = 58 m

I = 1 / (4π 582) = 2.4 x 10-5 W/m2 or 74 dBs

t030 m

t1

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� Three key pieces of information

� Time of echo

� Intensity of echo

� Frequency of echo

Plus prior knowledge of object being studied

� With modern technology (analog and digital) this can be done in real time.


Superposition

� Q: What happens when two waves “collide” ?

� A: They ADD together!� We say the waves are “superimposed”.

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Interference of Waves� 2D Surface Waves on Water

In phase sources separated by a distance d

d


Principle of superposition

� The superposition of 2 or more waves is called interference

Constructive interference:These two waves are in phase.Their crests are aligned.

Their superposition produces a

wave with amplitude 2a

Destructive interference:These two waves are out of

phase.The crests of one are aligned

with the troughs of the other.

Their superposition produces a

wave with zero amplitude

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Interference: space and time

� Is this a point of constructiveor destructive interference?

What do we need to do to make the sound from these two speakers interfere constructively?


Interference of SoundSound waves interfere, just like transverse waves do. The resulting wave (displacement, pressure) is the sum of the two (or more) waves you started with.

|||| 21 rrrrr −=∆

,...2,1,0

)21( 22

ceinterferen edestructiv Maximum

)(22

22

ceinterferen veconstructi Maximum

21

21

21

=

+=−+∆=∆

=−+∆=∆

=−+∆=∆

m

mr

mr

mr

πφφλπφ

λφφπλφ

πλ

πφφλπφ

])//(2cos[),( 22

2

2 φλπ +−= Ttrr

AtrD

])//(2cos[),( 11

1

1 φλπ +−= Ttrr

AtrD

r∆

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Example Interference� A speaker sits on a pedestal 2 m tall and emits a sine wave

at 343 Hz (the speed of sound in air is 343 m/s, so λ = 1m ). Only the direct sound wave and that which reflects off the ground at a position half-way between the speaker and the person (also 2 m tall) makes it to the persons ear.

� How close to the speaker can the person stand (A to D) so they hear a maximum sound intensity assuming there is no phase change at the ground (this is a bad assumption)?

The distances AD and BCD have equal transit times so the sound waves will be in phase. The only need is for AB = λ

t1

t0

t0

AB

A DC

d

h


Example Interference

� The geometry dictates everything else.

AB = λ AD = BC+CD = BC + (h2 + (d/2)2)½ = dAC = AB+BC = λ +BC = (h2 + d/22)½

Eliminating BC gives λ+d = 2 (h2 + d2/4)½

λ + 2λd + d2 = 4 h2 + d2

1 + 2d = 4 h2 / λ � d = 2 h2 / λ – ½= 7.5 m

t1

t0

t0

AB

A DC

7.5

4.253.25

Because the ground is more dense than air there will be a phase change of π and so we really should set AB to λ/2 or 0.5 m.

Page 8



Exercise Superposition

� Two continuous harmonic waves with the same frequency and amplitude but, at a certain time, have a phase difference of 170°are superimposed. Which of the follo wing best represents the resultant wave at this moment?

(A)

(E)

(D)

(C)

(B)

Original wave (the other has a different phase)


Wave motion at interfacesReflection of a Wave, Fixed End

� When the pulse reaches the support, the pulse moves back along the string in the opposite direction

� This is the reflection of the pulse

� The pulse is inverted

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Reflection of a Wave, Fixed EndAnimation


Reflection of a Wave, Free End

Animation

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Transmission of a Wave, Case 1

� When the boundary is intermediate between the last two extremes ( The right hand rope is massive or massless.) then part of the energy in the incident pulse is reflected and part is transmitted

� Some energy passes through the boundary

� Here µrhs > µlhs

Animation


Transmission of a Wave, Case 2

� Now assume a heavier string is attached to a light string

� Part of the pulse is reflected and part is transmitted� The reflected part is not inverted

Animation

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Standing waves� Two waves traveling in opposite direction interfere with each

other.

If the conditions are right, same k & ω, their interference generates a standing wave:

DRight(x,t)= a sin(kx-ωt) DLeft(x,t)= a sin(kx+ωt)A standing wave does not propagate in space, it “stands” in place.A standing wave has nodes and antinodes

D(x,t)= DL(x,t) + DR(x,t)

D(x,t)= 2a sin(kx) cos(ωt)The outer curve is the

amplitude function A(x) = 2a sin(kx)

when ωt = 2πn n = 0,1,2,…k = wave number = 2�/

�Nodes

Anti-nodes


Standing waves on a string

� Longest wavelength allowed is one half of a wave

Fundamental: λ/2 = L � λ = 2 L

,...3,2,1

2

=

==

m

fv

mL

mmλ

Recall v = f λ

Lvmfm 2

=Overtones m > 1

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Vibrating Strings- Superposition Principle

� Violin, viola, cello, string bass� Guitars� Ukuleles� Mandolins� Banjos

D(x,0)

An

tin

od

eD

(0,t

)


Standing waves in a pipeOpen end: Must be a displacement antinode (pressure minimum) Closed end: Must be a displacement node (pressure maximum) Blue curves are displacement oscillations. Red curves, pressure.

Fundamental: λ/2 λ/2 λ/4

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Standing waves in a pipe

,...3,2,1

2

2

=

=

=

m

Lvmf

mL

m

mλ

,...3,2,1

2

2

=

=

=

m

Lvmf

mL

m

mλ

,...5,3,1

4

4

=

=

=

m

Lvmf

mL

m

mλ


Combining Waves

Fourier Synthesis

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Lecture 29

•• AssignmentAssignment� HW12, Due Friday, May 8th

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