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by: Shanee Dinay

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# Week 7 Phys 5b Notes PHYS 5B

Shanee Dinay
UCSC
GPA 3.94

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## About this Document

Week 7 Physics 5b notes. Topics include Phase, Waves, Speaker Examples, Perfect Constructive, Perfect Destructive Interference, Double Split, Bright Fringes, Dark Fringes and more.
COURSE
Intro to Physics II
PROF.
A.Steinacker
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
Week 7, Physics 5b, physics 2, waves, Sound, interference, Fringes
KARMA
25 ?

## Popular in Physics 2

This 3 page Class Notes was uploaded by Shanee Dinay on Monday February 22, 2016. The Class Notes belongs to PHYS 5B at University of California - Santa Cruz taught by A.Steinacker in Fall 2015. Since its upload, it has received 10 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.

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Date Created: 02/22/16
Day 18 ­ 2/17/2016  Phys 5b    Midterms Back  ­ anyone with a score above 45 is in the running for an A  ­ that does not mean that if you have a 30 in the exam, then you have a chance  ­ provided that you show improvement  ­ she is interested in detailed steps  ­ she will drop the 1c) problem points  Phase, Waves  ϕ ϕ   1 2 y(x, t) = 2acos( Δϕ )sin(kx​agv​t ­ ϕ ​)avg​ Δϕ 2 A = | 2acos( 2 ) |  Δϕ = 2nπ → you get a perfectly constructive interference  Δϕ =  2π Δx + Δϕ   λ o A = |2a|  Δϕ = 2(n+1)π  → perfect destructive  A = 0  if Δϕ  = 0 identical speakers sources  o if Δϕ o  Δx  P.C.I 2πΔx  = 2nπ Δx = 2n  λ λ λ 2 P.D.I Δx = (2n + 1) 2 Example. Speakers  Two speakers, one meter away.   Δϕ o 2 f = 500 Hz a = 0.1mm  Δϕ A = |2a cos( 2 )|  Δx  =  x2− x 1= 1m  λ = cs  f λ = 343 m/ = .686 m  500 Hz Δϕ  =   λx + Δϕ  =o10.73 rad  10.73 A = | 2 • 0.1mm • cos 2 | = 0.121mm  Amplitude is larger than original amplitude (but not twice the original amplitude) so we have  partial constructive interference.  Destructive Interference when amplitude is between 0 and 1    Sound Waves, propagate away from source    Two Sources  Δϕ = 2λΔr + Δϕ o Δϕ  o 0  2π 2π Δϕ A λ (r2A ​ 1A​  λ (2λ  ­ 3λ ) = ­2π  perfect constructive interference  Δϕ = 2π (r​­ r​) =  2π(2.5λ  ­ 3λ) = ­π  A λ 2B ​ 1B​ λ perfect destructive interference  Δϕ A λπ(r2C ​ 1C​  2λ (3λ  ­ .5λ ) = 5π  perfect destructive  Where we have symmetry ­ antinode such as where Δr  = 0, Δr  = λ, Δr  = − λ, …   ­ we can draw nodal lines between the antinodal lines  Light  ­ light as a particle    Day 19 ­ 2/19/2016    Reading 34.3, 34.4, 35.7    Last Class Review: Narrow Beam of Light    Wave Front:  ← sources of spherical waves  ← planar front  Double Split    Δϕ  =   Δr  λ Δϕ Δϕ A = | 2a cos 2  | 2 = nπ  PCI → bright fringes  Δϕ = nπ Δϕ  =   Δr = 2nπ Δr = 2n   n = 0, 1, 2, …  2 λ 2 PDI ­ Dark Fringes  Δϕ 2 =(2n+1) 2 Δϕ = (2n + 1)π = λπΔr   λ Δr = (2n + 1)2 n = 1, 3, 5, …  Angles of Bright Fringes  λ θn= 2n 2d   Location of Bright Fringes  tanθn= yn y​n​ Ltanθ n yn​~ Lθ n= 2n  λL L 2d Location of Dark  λL y​dark​  = n(n + 1)  2d  Two consecutive fringes  Δy = y​ ­ y​  = 2(n+1) λL ­ 2n λL  n+1​ n​ 2d 2d Δy = λL   2d Electromagnetic Waves  2 I = CA​   2​ 2Δϕ 2​ 2Δϕ = C4a​ cos​  2 4Ca​ cos​  2 Δϕ A = | 2acos 2  |  a → amplitude of individual wave a  Δϕ = 2πΔr Δr = dsinθ  λ       =λπdsinθ   2​ π dy I(y) = 2Lo​os​ ( λ L Imin​0 Imax​4I​ o

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