4/7/2014
Ch. 11: waves
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Last time
Superposition of waves
Construction or Destruction
Reflection and Refraction
Demos
Examples
HITT quiz
Wave Summary
General form of travelling wave:
y = displacement of medium
{
perpendicular to propagation ⇒ transverse
parallel to propagation ⇒ longitudinal
Special case: periodic wave
Wave speed depends on properties of medium, e.g.
Intensity I ~ A2
Superposition of waves
The Principle of Superposition
Superposition Principle: When two or more waves overlap, the net disturbance
at any point is the sum of the individual disturbances due to each wave.
Two traveling wave
pulses: left pulse travels
right; right pulse travels
left.
displacements
(for general culture only!!)
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Example
Example: soln
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4/7/2014
Superposition of waves with same frequency,
amplitude and wavelength but different phase
Superposition of waves cont’d
constructive
destructive
Superposition of waves with same frequency
and wavelength but different phase
Interference
When two waves travel different
distances to reach the same point, the
phase difference is determined by:
d1 − d 2
λ
phase difference ∆φ =
∆φ = k∆d = 2πn, Constructive
2πn
∆d =
= nλ , n = 0,±1, ± 2,⋅ ⋅
k
Example Problem: Superposition
S1
S2
S3
S4
∆φ13 = k∆d13 =
2π∆d13
λ
=
2π ( d1 − d 2 )
λ
phase difference ∆φ
=
2π
2π
= k ( d1 − d 2 )
1
∆φ = k∆d = 2π (n + ), Destructive
2
2π
1
1
∆d =
(n + ) = (n + )λ , n = 0,±1, ±102,⋅ ⋅
k
2
2
Example Problem: Superposition
P
d
d
d
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 λ 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 = λ/4?
Answer: Zero
d1 = x + 3d
=
2π 2d
λ
=
2π (λ / 2)
λ
=π
d 2 = x + 2d
2π∆d 24 2π 2d 2π (λ / 2)
=π
d3 = x + d ∆φ24 = k∆d 24 = λ = λ =
λ
d4 = x
Destructive
Destructive
• 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?
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 = πr to reach the
point B.
∆d = d 2 − d1 = (π − 2)r = (n + 12 )λ
r=
PHY 2053
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Point A Point B
n = 0,±1,±2, L
Destructive
(n + 12 )λ
(40cm)
λ
→
=
≈ 17.5cm
(π − 2) min 2(π − 2) 2(π − 2)
PHY 2053
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Reflection
Reflection and Refraction
When you have a wave that travels
from a “low density” medium to a “high
density” medium, the reflected wave
pulse will be inverted.
At an abrupt boundary between two media, a reflection will occur. A portion of
the incident wave will be reflected backward from the boundary.
The frequency of the reflected wave
remains the same.
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Traveling Waves: Refraction
Traveling Waves: Refraction
• Refraction: Incident Angle θ1
• Refraction: Zero Incident Angle
(speed of propagation is the frequency times the wavelength)
v1 = f1λ1
v2 = f 2λ2
∆t =
λ1
v2
Medium 1
Speed v1
Medium 2
Speed v 2
v1
=
v 2 λ1
v1
λ1
v1
f1 = f 2
Crest A
λ2
v1 = fλ1
sin θ 2 =
sin θ 1
λ1
λ2
(Law of Refraction)
f1 = f 2
v 2 = fλ 2
sin θ 1 =
v2
(distance traveled by crest A in time ∆t)
v2
=
=
λ1
L
λ2
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λ1
λ1
v1
λ2
θ2
L
λ2
L
Medium 2
Speed v2
(incident angle)
sin θ 2
θ1
Medium 1
Speed v1
(refracted angle)
sin θ 1 sin θ 2
=
v1
v2
(frequency is the same)
PHY 2053
v1
(speed of propagation is the frequency times the wavelength)
λ1
v2 < v1
λ2 = v2 ∆ t =
λ2
Crest B
(time for crest B to reach medium 2)
v1
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v2 < v1
v2
(Law of Refraction)
(Law of Refraction)
(Law of Refraction)
PHY 2053
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