Transcript 2053_Lecture_11-14-13

Traveling Waves: Superposition
• Wave Superposition:
y-axis
y
y1(x,t) A1
1
y1 ( x, t )  A1 sin( 1 )  A1 sin( k1x  1t )
A
f
y-axis
y
y2(x,t) A2
2
y2 ( x, t )  A2 sin( 2 )  A2 sin( k2 x  2t )
A
f
Add the two waves together (superposition of wave 1 and wave 2) as follows:
y12 ( x, t )  y1 ( x, t )  y2 ( x, t )  A1 sin( 1 )  A2 sin(  2 )
 A1 sin( k1 x  1t )  A2 sin( k2 x  2t )
R. Field 11/14/2013
University of Florida
PHY 2053
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Traveling Waves: Superposition
• Wave Superposition:
y1 ( x, t )  A1 sin( 1 )  A1 sin( k1x  1t )
y2 ( x, t )  A2 sin( 2 )  A2 sin( k2 x  2t )
Wave 2
y-axis
Superposition!
y-axis
Wave 1
A2
A12 A2
2
1
A1
1
A1
2
y12 ( x, t )  y1 ( x, t )  y2 ( x, t )  A1 sin( 1 )  A2 sin(  2 )
 A1 sin( k1 x  1t )  A2 sin( k2 x  2t )
The intensity of the new wave is proportional to A12 squared!
R. Field 11/14/2013
University of Florida
PHY 2053
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Traveling Waves: Superposition
• Wave Superposition: Consider two waves with the same amplitude,
frequency, and wavelength but with an overall phase difference of D = f.
y1 ( x, t )  A sin( 1 )  A sin( kx  t  f / 2)
y2 ( x, t )  A sin( 2 )  A sin( kx  t  f / 2)
sinA+sinB = 2sin[(A+B)/2]cos[(A-B)/2]
D  2  1  f
y-axis
A
D
A
y12 ( x, t )  y1 ( x, t )  y2 ( x, t )
 A sin( kx  t  f / 2)  A sin( kx  t  f / 2)
 2 A cos f sin( kx  t )
Superposition!
y12 ( x, t )  A12 sin( kx  t )
A12  2 A cos f  2 A cos D
New amplitude!
I  I1  I 2  A 2
I12  A122  4 A2 cos 2 f
New intensity!
I12  4 I cos 2 f  4 I cos 2 D
R. Field 11/14/2013
University of Florida
PHY 2053
Page 3
Traveling Waves: Interference
• Maximal Constructive Interference: Consider two waves with the same
amplitude, frequency, and wavelength but with an overall phase difference
of D = 2pn, where n = 0, ±1, ±2,…
D  2  1  2pn
y-axis
A
A
n  0,1,2,
y12 ( x, t )  2 Asin( kx  t )
I1  I 2  I
I12  4I
Max Constructive!
• Maximal Destructive Interference: Consider two waves with the same
amplitude, frequency, and wavelength but with an overall phase difference of
D = p+2pn, where n = 0, ±1, ±2,…
D  2  1  p  2pn
y-axis
D
A
n  0,1,2, 
y12 ( x, t )  0
A
I1  I 2  I
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PHY 2053
I12  0
Max Destructive!
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Example Problem: Superposition
• Two traveling pressure waves (wave A and wave B) have the same frequency and
wavelength. The waves are superimposed upon each other. The amplitude of the
resulting wave (wave C) is 13 kPa. If the amplitude of wave A is 12 kPa and the phase
difference between wave B and wave A is fB – fA = 90o, what is the amplitude of wave B
and the magnitude of the phase difference between wave A and wave C, respectively?
Answer: 5 kPa, 22.62o
cos  AC 
AB
AA 12

 0.923
AC 13
AC
AB
 AC  22.62
AC
90o
AA
AA
AA2  AB2  AC2
AB  AC2  AA2  (13kPa) 2  (12kPa) 2  5kPa
R. Field 11/14/2013
University of Florida
PHY 2053
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Traveling Waves: Superposition
• Lateral Phase Shift: Consider two waves with the same amplitude,
frequency, and wavelength that are in phase at x = 0.
D  2  1  k (d2  d1 )  kDd
1  kd1  t
2
A
= -t
Wave 1 distance d1
A
A
1
2  kd2  t
x=0
Wave 2 distance d2
A
= -t
x=0
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D  kDd  2pn Max Constructive
2pn
Dd 
 n n  0,1,2,
k
Max Destructive
D  kDd  p  2pn
(p  2pn)
Dd 
 (n  12 ) n  0,1,2,
k
PHY 2053
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Examples: Superposition
Dd = /2 max destructive
Dd =  max constructive
Wave Superposition
2.0
1.5
1.0
0.5
0.0
-0.5 0
-1.0
-1.5
-2.0
Wave Superposition
1.0
0.5
0.0
1
2
3
4
5
6
7
0
8
1
2
3
4
5
6
7
-0.5
-1.0
ysum = y1 + y2
ysum = y1 + y2
kx (radians)
Dd = /4
kx (radians)
Wave Superposition
1.5
1.0
0.5
0.0
-0.5
0
1
2
3
4
5
6
7
8
-1.0
-1.5
ysum = y1 + y2
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kx (radians)
PHY 2053
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8
Example Problem: Superposition
S1
S2
S4
S3
P
x
• The figure shows four isotropic point sources of sound that are uniformly spaced on
the x-axis. The sources emit sound at the same wavelength  and the same amplitude
A, and they emit in phase. A point P is shown on the x-axis. Assume that as the sound
waves travel to the point P, the decrease in their amplitude is negligible. What is the
amplitude of the net wave at P if d = /4?
d
d
d
3
Answer: Zero
d1  x  3d
d 2  x  2d
d3  x  d
D 34  kDd 34 
D 23  kDd 23 
d4  x
D12  kDd12 
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2pDd 34

2pDd 23

2pDd12




2pd

2pd

2pd

PHY 2053



2p ( / 4)

2p ( / 4)

2p ( / 4)


p


2
4
2
p
2
1
Max Destructive
p
2
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Example Problem: Superposition
• Sound with a 40 cm wavelength travels rightward from a
source and through a tube that consists of a straight portion
and a half-circle as shown in the figure. Part of the sound
wave travels through the half-circle and then rejoins the rest of
the wave, which goes directly through the straight portion.
This rejoining results in interference. What is the smallest
radius r that results in an intensity minimum at the detector?
Point A Point B
Answer: 17.5 cm
At point A the waves have the same amplitude, wavelength, and frequency and are in phase.
Wave 1 travels a distance d1 = 2r to reach the point B, while wave 2 travels a distance d2 = pr to
reach the point B.
Dd  d 2  d1  (p  2)r  (n  12 )
n  0,1,2,
Max Destructive
(n  12 )

(40cm)
r
min


 17.5cm
(p  2)
2(p  2) 2(p  2)
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University of Florida
PHY 2053
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