Transcript Physics 207: Lecture 2 Notes

Lecture 29
Goals:
• Chapter 20
•
 Work with a few important characteristics of sound waves.
(e.g., Doppler effect)
Chapter 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.
• Assignment
 HW13, Due Friday, May 7h
 Thursday, Finish up, begin review for final
Physics 207: Lecture 29, Pg 1
Doppler effect, moving sources/receivers
Physics 207: Lecture 29, Pg 2
Doppler effect, moving sources/receivers
 If the source of sound is moving
 Toward the observer 
 seems smaller
 Away from observer 
 seems larger
f observer
f source

vs
1 v
f observer
f source

vs
1 v
 If the observer is moving
 Toward the source  f observer
 seems smaller
 Away from source 
 seems larger
f observer
 vo 
 1   f source
v

 vo 
 1   f source
v  Doppler Example Audio

Doppler Example Visual
Physics 207: Lecture 29, Pg 3
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  = 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
30 m
A
Physics 207: Lecture 29, Pg 4
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?
f
f observer 
At the wall: fwall = 340 / (1-20/340) = 361 Hz
source
vs
1 v
Wall becomes “source” for the subsequent part
 vo 
f observer  1   f source
v

At the speaker f ’ = fwall (1+ 20/340) = 382 Hz
t0
t1
30 m
Physics 207: Lecture 29, Pg 5
Example Sound Intensity

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 Dt + vsound Dt = 2 x 30 m = 60 m
Dt = 60 / (340+20) = 0.17 s  dsound = 340 * 0.17 m = 58 m
I = 1 / (4p 582) = 2.4 x 10-5 W/m2 or 74 dBs
t0
t1
30 m
Physics 207: Lecture 29, Pg 6
Doppler effect, moving sources/receivers
 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.
Physics 207: Lecture 29, Pg 7
Ch. 21: Wave Superposition
 Q: What happens when two waves “collide” ?
 A: They ADD together!
 We say the waves are “superimposed”.
Physics 207: Lecture 29, Pg 8
Interference of Waves
 2D Surface Waves on Water
In phase sources separated
by a distance d
d
Physics 207: Lecture 29, Pg 9
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
Physics 207: Lecture 29, Pg 10
Interference: space and time
 Is this a point of constructive
or destructive interference?
What do we need to do to
make the sound from these
two speakers interfere
constructively?
Physics 207: Lecture 29, Pg 11
Interference of Sound
Sound waves interfere, just like transverse waves do. The
resulting wave (displacement, pressure) is the sum of the two (or
more) waves you started with.
A
D(r1 , t )  cos[ 2p (r1 /   t / T )  1 ]
r1
D(r2 , t ) 
A
cos[ 2p (r2 /   t / T )  2 ]
r2
Path Difference


Dr | r1 |  | r2 |
Dr
Physics 207: Lecture 29, Pg 12
Interference of Sound
Sound waves interfere, just like transverse waves do. The
resulting wave (displacement, pressure) is the sum of the two (or
more) waves you started with.
A
D(r2 , t )  cos[ 2p (r2 /   t / T )  2 ]


r2
Dr | r1 |  | r2 |
A
D(r1 , t )  cos[ 2p (r1 /   t / T )  1 ]
r1
Maximum constructi ve interferen ce
D  2p Dr  1  2  2p m



D  Dr 
(1  2 )  m
2p
2p
Maximum destructiv e interferen ce
D  2p Dr  1  2  2p (m  1 )
2

m  0,1,2,...
Dr
Physics 207: Lecture 29, Pg 13
Example Interference


A speaker sits on a pedestal 2 m tall and emits a sine wave
at 340 Hz (the speed of sound in air is 340 m/s, so  = 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)?
t1
t0
d
A
t0
B
D
h
C
The distances AD and BCD have equal transit times so the
sound waves will be in phase. The only need is for AB = 
Physics 207: Lecture 29, Pg 14
Example Interference

The geometry dictates everything else.
AB = 
AD = BC+CD = BC + (h2 + (d/2)2)½ = d
AC = AB+BC =  +BC = (h2 + d/22)½
Eliminating BC gives
+d = 2 (h2 + d2/4)½
 + 2d + d2 = 4 h2 + d2
1 + 2d = 4 h2 /   d = 2 h2 /  – ½
= 7.5 m
t1
t0
7.5
A
t0
B
3.25 C
4.25
D
Because the ground is more dense than air there will be a phase
change of p and so we really should set AB to /2 or 0.5 m.
Physics 207: Lecture 29, Pg 15
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 following
best represents the resultant wave at this moment?
Original wave
(the other has a different phase)
(A)
(B)
(D)
(C)
(E)
Physics 207: Lecture 29, Pg 16
Wave motion at interfaces
Reflection 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
Physics 207: Lecture 29, Pg 17
Animation
Reflection of a Wave, Fixed End
Physics 207: Lecture 29, Pg 18
Reflection of a Wave, Free End
Animation
Physics 207: Lecture 29, Pg 19
Standing waves
 Two waves traveling in opposite direction interfere with each
other.
If the conditions are right, same k & w, their interference
generates a standing wave:
DRight(x,t)= a sin(kx-wt) DLeft(x,t)= a sin(kx+wt)
A standing wave does not propagate in space, it “stands” in place.
A standing wave has nodes and antinodes
Anti-nodes
D(x,t)= DL(x,t) + DR(x,t)
D(x,t)= 2a sin(kx) cos(wt)
The outer curve is the
amplitude function
A(x) = 2a sin(kx)
when wt = 2pn n = 0,1,2,…
k = wave number = 2π/λ
Physics 207:Nodes
Lecture 29, Pg 22
Standing waves on a string
 Longest wavelength allowed is
one half of a wave
Fundamental: /2 = L   = 2 L
v
2
L
m 

m
fm
m  1,2,3,...
Recall v = f 
v
fm  m
2L
Overtones m > 1
Physics 207: Lecture 29, Pg 23
Vibrating Strings- Superposition Principle
 Violin, viola, cello, string bass
D(x,0)
 Guitars
 Ukuleles
 Mandolins
Antinode D(0,t)
 Banjos
Physics 207: Lecture 29, Pg 24
Standing waves in a pipe
Open 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:
/2
/2
/4
Physics 207: Lecture 29, Pg 25
Standing waves in a pipe
m  2 L
m  2 L
m  4 L
fm  m v
2L
m  1,2,3,...
fm  m v
2L
m  1,2,3,...
fm  m v
4L
m  1,3,5,...
m
m
m
Physics 207: Lecture 29, Pg 26
Combining Waves
Fourier Synthesis
Physics 207: Lecture 29, Pg 27
Organ Pipe Example
A 0.9 m organ pipe (open at both ends) is measured to have
it’s first harmonic (i.e., its fundamental) at a frequency of
382 Hz. What is the speed of sound (refers to energy
transfer) in this pipe?
L=0.9 m
f = 382 Hz and f  = v with  = 2 L / m (m = 1)
v = 382 x 2(0.9) m  v = 687 m/s
Physics 207: Lecture 29, Pg 28
Standing Waves
 What happens to the fundamental frequency of a pipe, if
the air (v =300 m/s) is replaced by helium (v = 900 m/s)?
Recall: f  = v
(A) Increases
(B) Same
(C) Decreases
Physics 207: Lecture 29, Pg 29
Lecture 29
• Assignment
 HW13, Due Friday, May 7th
Physics 207: Lecture 29, Pg 30