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Types of Waveforms Flipbook PDF
Types of Waveforms • Harmonic (aka sinusoidal) – Special case of periodic x displacement travels in time (Longitudinal a
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Types of Waveforms • Pulse: not periodic – drops to zero on both sides
displacement travels in time
– f(x)
x
• Continuous (or periodic) wave: displacement travels in time x λ June 27, 2011
6
Types of Waveforms • Harmonic (aka sinusoidal) – Special case of periodic
(Longitudinal and transverse)
displacement travels in time A
x
λ
– Wave speed: v =
June 27, 2011
λ T
= λf 7
Mathematical Description • Pulse: f (x ) • Moving pulse:
– “+” direction f ( x − vt ) – “-”direction f ( x + vt ) travels in time f(x) f(x-vt)
x
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Simple Harmonic Wave displacement
• Position at t=0: –
2πx y ( x,0 ) = A cos = A cos(kx ) λ
A
x λ
• Travelling to the right with speed, v: y ( x, t ) = y ( x − vt ) = A cos k ( x − vt ) = – A cos(kx − ωt ) ω = 2πf =
2π T
• Angular frequency---> 2π k = • Wave number---------> λ ω v = • Velocity-----------------> k = λf June 27, 2011
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displacement A
t T
Waves on a String under tension • Small amplitude displacements • transverse • Given mass density, μ, tension, F Speed: v =
F
– µ – Dimensions: [F]/[μ]=[v]
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v
Average Power (waves propagate energy) v v v v • P=dE/dt= F ⋅ u (W = F ⋅ x )
∂y – Speed of medium u = = ωA sin (kx − ωt ) ∂t
• Power fluctuates throughout the wave (book) – P = FωkA2 sin 2 (kx − ωt )
• Interested in average power 1 – P = FωkA2 2
1 2 2 P = µω Av • Waves on a string 2
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Wave Intensity • Intensity= average energy/time/area (W/m^2) – Area perpendicular to propagation direction
• Wavefront – surface of constant phase – Plane wavefronts (plane wave) – constant intensity – Spherical wavefront (spherical wave) – intensity decreases as distance from source increases P P I= = A 4πr 2 June 27, 2011
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Superposition • Wave interference = algebraic sum of displacements
Wave1 Wave2
Constructive (in phase) ∆φ = 2πn
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Destructive (180 degree out of phase) ∆φ = nπ
Beats • Interference of 2 waves w/slightly different frequencies • At a fixed position, x=0: y (t ) = A cos ω1t + A cos ω2t ω1 − ω2 ω1 + ω2 = 2 A cos t cos t 2 2
“Amplitude” June 27, 2011
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Dispersion • Wave frequency (speed) depends on wavelength v = λf (λ )
• Waves on the surface λg v= 2π
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Wave Equation • All equations of the form f ( x ± vt ) satisfy ∂2 f 1 ∂2 f = 2 2 2 ∂x v ∂t
• i.e. displacement of waves on a string y ( x, t ) = A cos(kx ± ωt )
∂2 y µ ∂2 y = 2 ∂x F ∂t 2 June 27, 2011
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Sound Waves In Gas • Longitudinal pressure wave pressure γP • Speed of wave v = ρ
density
• Coefficient, γ, depends on nature of gas – Ideal γ = 5 / 3 – Diatomic γ = 7 / 3
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Sound waves in Liquid/Solid • Speed of sound B γP v= v= (in gas, ) ρ ρ • Bulk modulus (wave speed depends on direction) ∆P B=−
∆V / V
– Units of pressure – Measure of “stiffness” (large B, hard to compress)
• longitudinal & transverse June 27, 2011
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Sound Intensity P I= Area
• Recall for a string, • In terms of pressure,
1 2 2 P = µω Av , 2
Power = Force ⋅ velocity = (∆P )( Area )u Area
• Pressure, displacement vary sinusoidally ∆P = (∆P0 ) cos(kx − ωt )
s = − s0 sin (kx − ωt ) 2 ( ∆P0 ) • See pg 419-420 I =
1 2 = ρω 2 s0 v 2 ρv 2
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Decibels (dB) • Human ear can detect a wide range of frequencies log scale! • Decibel unit defined as β = 10 log(I I 0 ) where I 0 = 10−12W / m 2 is threshold of hearing at 1kHz • >40dB, perceived loudness doubles for every 10dB increase June 27, 2011
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Standing waves Fundamental, m=1
• String (L) clamped tightly at both ends • No propagation (superposition of waves propagating in opposite directions • Allowed modes mode number mλ1 L=
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First harmonic
Second harmonic
node
2 20
antinode
Standing waves • String clamped at one • String open at both end ends
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Standing waves are everywhere!
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*GRADES – will be curved, but only in your favor
Review - waves ω
• v = λf = k (different from velocity of medium) ∂ f 1 ∂ f f ( x m vt ) • Wave equation – ∂x = v ∂t A cos(kx − ωt ) – Wave on a string v = F µ 2
2
2
2
2
– Sound in gas v = γP ρ – Sound in solid/liquid v = B ρ (bulk modulus B = − ∆P (∆V /V ) )
k=
• Longitudinal and/or transverse • Superposition (i.e. beats) • Dispersion (wave packets spread out) June 28, 2011
1
2π
λ
ω=
2π T
ω1 + ω2 2
incident wave
reflected wave
• Propagation through bounded medium • Superposition of 2 waves travelling in opposite directions • 3 bounded scenarios
mλ L= 2 mλ L= 4 (m + 1 2)λ L= 2
– Closed on both – Open on both – open on one
node
antinode
f 0 = v λmax
harmonics
Review – Standing waves
amplitude vs instantaneous amplitude June 28, 2011
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Review - Intensity 1 • Average power (string)– P = µω 2 A2v 2
• “wavefront” – surfaces of constant phase • Plane/spherical wave – plane/spherical wavefronts • Intensity – characterizes energy flow in >1D – 3D
I=
P P = A 4πr 2
(2D
I2D
• Intensity of sound
decibels β = 10 log(I I 0 )
2 ( ∆P0 ) I=
1 2 = ρω 2 s0 v 2 ρv 2
June 28, 2011
P P = = ) l 2πr
3
Doppler effect • Wave propagates symmetrically in every direction from a point source at rest • When source/observer moves, the observed frequency and/or wavelength is shifted • Let u be speed of source or observer June 28, 2011
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Doppler: source moving
u
f ′ = f (1 + u / v )
• Source reference frame:: travel λ in one period, T • During T, source covers distance of uT • Observed wavelength: λ ′ = λ m uT λ ′ = λ (1 m u / v ) • frequency f ′ = f (1 m u / v ) June 28, 2011
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f ′ = f (1 − u / v )
Doppler Shift: Moving Observer • • • •
Shift in frequency only, wavelength does not change Speed observed = v+u Observed period T ′ = λ
v u
v+u
• Observed frequency shift
f ′ = f (1 ± u / v )
(negative sign means observer moving AWAY) June 28, 2011
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EM waves
June 28, 2011
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4 laws of EM (up to now) Gauss
no magnetic v v ∇ ⋅ E = ρ ε 0 monopoles v v v v v v ∇ ⋅ B = 0 Faraday ∫∫∫ ∇ ⋅ Fdτ = ∫∫ F ⋅ dA v v v v v v v v ∇ × E = − ∂B ∂t ∇ × F ⋅ dS = ∫ F ⋅ d l ∫∫ v v v ∇ × B = µ0 J Ampere
dB/dt
Changing E-fields induce B-fields?
E June 28, 2011
v v ∫ E ⋅ dA = qencl ε 0 v v ∫ B ⋅ dA = 0 Lenz’s law v v ∫ E ⋅ d l = − ∂Φ B ∂t v v ∫ B ⋅ d l = µ0 I encl
8
Modify Ampere’s law • Ambiguity in ampere’s law v v ∫ B ⋅ d l = µ0 I encl v v ∫l1B ⋅ d l = µ0 I v v ∫ B ⋅ d l = 0!
l1
l2
l2
• Displacement current I D = ε 0 ∂Φ E ∂t
Amperean loop
v v ∫ B ⋅ d l = µ0 I encl + µ0ε 0 ∂Φ E ∂t June 28, 2011
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Maxwell’s Equations Simplest scenario: in a vacuum - complete v v Gauss ∫ E ⋅ dA = qencl ε 0 v v ∫ B ⋅ dA = 0 v v ∫ E ⋅ d l = − ∂Φ B ∂t Faraday v v ∫ B ⋅ d l = µ0 I encl + µ0ε 0 ∂Φ E ∂t Ampere (complete)
Electric and Magnetic fields on equal footing June 28, 2011
v v ∫ E ⋅ dA = 0 v v ∫ B ⋅ dA = 0 v v ∫ E ⋅ d l = − ∂Φ B ∂t v v ∫ B ⋅ d l = µ0ε 0 ∂Φ E ∂t Electromagnetic waves!
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EM Waves • Coupled differential equations v v Take the curl of both sides ∇⋅E = 0 v v ∇⋅B = 0 v v v ∇ × E = − ∂B ∂t v v v ∇ × B = µ0ε 0 ∂E ∂t
v v v v v v0 2v ∇ × (∇ × E ) = ∇(∇ ⋅ E ) − ∇ E v 2 v v v ∂ ∂ ∂ E − (∇ × B ) = − (µ0ε 0 ∂E ∂t ) = − µ0ε 0 2 ∂t ∂t ∂t ∂2 f 1 ∂2 f = 2 2 2 ∂x c ∂t
• In one dimension v v 2 2 ∂ E ∂ E = µ0ε 0 2 2 ∂x ∂t June 28, 2011
c= 10
1
µ0ε 0
= 3.0 × 10 m / s 8
EM plane Wave • E,B harmonic waves, oscillate in perpendicular directions • We use x for the propagation direction v E ( x, t ) = E p cos(kx − ωt ) ˆj Amplitudes related speed of wave v B ( x, t ) = B p cos(kx − ωt )kˆ y
E
c =ω k
E
In phase
h
Δx
Δx
h
z B June 28, 2011
E p = cB p
11
x
Electromagnetic Spectrum • Continuum of allowed wavelengths (frequencies) as long as c = λf
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Polarization • Specifies direction of E field v E ( x, t ) = E p cos(kx − ωt )iˆ – Polarization is perpendicular to B, v
• Unpolarized: superposition of waves with a random orientation of polarizations – i.e. visible light from sun, light bulb
• Unpolarized becomes polarized… – Reflection from surfaces (partial) – POLARIZER (crystal with preferred direction of transmission, called transmission axis) June 29, 2011
1
Polarizers
nothing
E0 cos θ E0
• Polarizers reorient a component of the incident wave – The output light is completely polarized along the polarizer axis – Polarizers change the light – Polarizers are not filters • Filters can only decrease (or leave unchanged) the light • Filters cannot add light no filter can add red light Filters: white light June 29, 2011
2
blue light
?
nothing
Polarizers – Law of Malus • Only component of E along transmission axis will pass through unattenuated
E f = E0 cos θ0
• Intensity (S) varies as amplitude squared
S = S0 cos θ0 2
• For unpolarized light, average over ever direction to get emerging intensity
Sunpolarized June 29, 2011
3
1 = S0 2
Producing EM Waves • Changing B E; changing EB – Accelerated electrical charges
• If motion of charges is periodic, EM with that frequency • Systems are “good” transmitter/receivers if size of system d~λ • Directional antenna (TV bunny) • Approximate plane wave June 29, 2011
4
Energy in EM Waves • Energy density (energy/volume)
(
v2 v2 1 u = ε 0 E + B µ0 2
)
A
• Want intensity (energy/time/area) dU = udVol = uAdx • dU dx = uA dx c dt v2 v2 1 dU c S= = ε 0 E + B µ0 A dt 2 v2 rewrite S = ε 0c E v v
)
(
• Can
S= June 29, 2011
EB
µ0 5
S=
v2 cB
µ0
dx
c
(E
p
= cB p )
v E ( x, t ) = E p cos(kx − ωt ) ˆj v B ( x, t ) = B p cos(kx − ωt )kˆ
Poynting Vector • Energy/area/time v v
v E×B S=
y
µ0
• Points in direction of propagation (RHR) v z • Wavevector k v S
v E v B June 29, 2011
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x
Average intensity (magnitude) • Poynting vector magnitude in terms of E v S =
v v EB
µ0
=
E p Bp
µ0
cos (kx − ωt )
• Average value cos^2=1/2 2 S=
June 29, 2011
S=
E p Bp 2 µ0
S= 7
ε 0cE p 2 v 2 cB p 2 µ0
2
2 lasers emit light of the same color, but the E field in laser 1 is 2x that of laser 2. compare their: B-fields intensities wavelengths
Waves from point source • Spherical waves Power S= 4πr 2
• Energy spreads in space over time, amplitude does not change • Field strength falls off as 1/r – EM waves dominate in all but the immediate vicinity of accelerating charges
June 29, 2011
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Wave Momentum • EM wave energy and momentum related by p=
U c
– If average energy/t/area= S
• Average momentum/t/area=> v v • Newton’s second law F = dp dt • Force/Area = Pressure,
S c
– RADIATION PRESSURE
• EM wave delivers 2x pressure to object on reflection June 29, 2011
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Ch 17 Review (Sound & more Waves) • Speed of (longitudinal) wave is v=
– Gases – Liq/Sol
v=
γP ρ
(in air, 340 m/s)
B
B=−
ρ
∆P ∆V / V
• Intensity –
2 ( ∆P0 ) I=
2 ρv
=
1 ρω 2 s0 2v 2
• Decibels
– β = 10 log(I I 0 )
• Standing waves • Doppler effect
– Moving source f ′ = f (1 ± u / v ) – Moving Observer f ′ = f (1 ± u / v )
July 5, 2011
1
Examples • Ch 17-22 A 1-dB increase in sound level is about the minimum change the human ear can perceive. By what factor is the sound intensity increased for this dB
• Ch 17 46 Estimate the fundamental frequency of the human vocal tract by assuming it to be a cylinder 15 cm long that is closed on one end.
change?
L = λmax 4
∆β = β 2 − β1 = 10 log(I 2 I1 )
I 2 I1 = 10 I 2 I1 = 10
1 10
∆β
L = 3λ 4
10
= 1.26dB
L = 5λ 4
v = λmax f 0 = 340m / s f0 = v
July 5, 2011
2
4L
=
(340m / s )
1m 415cm × 100cm
= 570 Hz
Ch 34 Review (EM Waves) • Maxwell’s Complete Equations v v ∫ E ⋅ dA = qencl ε 0 v v B ∫ ⋅ dA = 0 v v Displacement E ⋅ d l = − ∂ Φ ∂ t B ∫ current v v ∫ B ⋅ d l = µ0 I encl + µ0 (ε 0 ∂Φ E ∂t )
v v ∫ E ⋅ dA = 0 v v B ∫ ⋅ dA = 0 v v E ∫ ⋅ d l = − ∂Φ B ∂t v v ∫ B ⋅ d l = µ0ε 0 ∂Φ E ∂t
v v ˆ E ( x, t ) = E0 cos(kx − ωt ) j B ( x, t ) = B0 cos(kx − ωt )kˆ polarization
• Intensity:
– Poynting vector
v v v E×B S=
• S → S cos θ (S → S 2) U p = • EM momentum c 2
0
0
– Rad Press July 5, 2011
0
S c 3
0
µ0
v v 2 ∂ E 1 ∂ E = 2 2 2 c ∂t ∂x 2
c=
E0 = cB0
1
µ0ε 0
=
ω k
Examples • Ch 34-64 • Ch 34-6 Find the angle between 2 polarizers if An E-field points into the page and upolarized light incident on the pair occupies a circular region of radius emerges with 10% of its incident 1m. There is B-field forming CCW intensity. loops. The B-field strength 50cm from After 1st polarizer… center is 2.0-uT. (a) What is the rate of 1 change of E-field? (b) Is the E-field S1 = S0 2 increasing or decreasing? B
v v dE/dt=? ∫ B ⋅ d l = µ0ε 0 ∂Φ E ∂t
xxx xxxxx x x x xr x x x Rx x x x x xxx
B (2πr ) = µ0ε 0 (∂E ∂t )πr 2 B (@ r ) = µ0ε 0 r 2 (∂E ∂t )
(∂E ∂t ) = 2 Bc 2
r
CCW B dE/dt out of page by RHR! So strength of E - decreasing July 5, 2011
4
After 2nd polarizer
S2 = S1 cos2 θ0
= (S0 2 ) cos2 θ0
S2 cos2 θ0 = 0.10 = 2 S0