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by: Shanee Dinay

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# Week 6 Phys 5b Notes PHYS 5B

Shanee Dinay
UCSC
GPA 3.94

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Notes for Physics 5b Intro to Physics 2 Notes. Topics include Chapter 15 and 16, Standing Waves, Transverse Wave, String Tension, Pulse, Wave Speed, Gas Behavior, Linearization Process, Wave Intens...
COURSE
Intro to Physics II
PROF.
A.Steinacker
TYPE
Class Notes
PAGES
8
WORDS
CONCEPTS
Physics 5b, Physics, waves, Pulse, Wave Speed, Wave Intensity
KARMA
25 ?

## Popular in Physics 2

This 8 page Class Notes was uploaded by Shanee Dinay on Monday February 15, 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: 02/15/16
Day 15 ­ 2/8/2016  Phys 5b    HW #1  ­ asked to calculate how long will it take for the displacement to decrease one percent to  its original value, referring to the exponential envelope. Pretend as if the sinusoidal didn’t  exist  Reading: 15.2, 15.5, 15.7, 16.1, 16.2  Review: Standing Waves  ­ by adding up two waves of the same amplitude, wavelength, and frequency, but  propagating in opposite direction, y​ 1​, t) = Asin(kx ­ wt) and y​ 2​x, t) = Asin(kx + wt), we  obtained a standing wave, and we have shown that its mode of vibration depend on the  boundary conditions as follows    Closed Closed Standing Wave Pattern  λ = 2L n = 1, 2, 3, …  n n P n = vw = n vw → T​  =  = 12L  λn 2L n​ fn nvw n↑ → T↓  → v​ p​↑  y(x, t) = Asin(kx ­ wt)  v​ =  dy = wA cos(kx ­ wt)  p​ dt   the blue is the sine graph  the purple is the velocity of the graph above it  ­ negative velocity, it is moving downward    Tension force on the rope:    forces are as follows:    An upwards pull up:      We want to use these ideas to derive the wave equation for the transverse wave:    F​ = T​  ­ T​ = T​ sin(α + aα) ­ T​ sin(α)  y​ 2y​ 1y​ 2​ 1​ T2x​= T​1x​→ T​2​= T​1   T2​os(α + aα) = T​ 1​os(α)  sin(α + a α)  → α + aα    2 dm d y = T​(α + aα) ­ T​ α  dt s​ c​ d y Ts​aα = dm dt   Linear density of our string  dm linear density =  dl  d y Ts​α = μdx  dt   2 dα = μ d2y α =  dy  d2 Tsd2 dx d y = μ d y → [T 2] =  N  =  kg • m/ =  m2  dx2 Tsdt2 μ kg /m kg / m s2 d y 1 d y =  dx2= v t2   v​ =  Ts  w​ √ μ   n = 2  vw fn​ n  2L = f  Ts​↑ 4x L = constant, f = constant  If we increase the tension, the speed of the wave is going to stay the same  v​ =  4Tso = 2v​  w0​ √ μ   w0 vw f = n 2L  n↑ → v​ w​  and we get a mode of 1    L = 2.5m  y(x, t_ = 2Acos(wt)sin(kx​ n​ xn​→ y(x​n​ t) = 0  sin(kx​) = 0  n​ kxn​= nπ  2π λ Location of mode λxn= nπ → x​ n​= n 2 λ λ λ xn+1​x​n​= (n+1) 2­ n 2=   2 λ 2  = 0.5m → λ  =  1m        Left column shows incoming pulse and ghost pulse  Right column shows us what the pulse actually looks like   as time goes on, the pulse flips with the interaction with  the wall    Keep temperature constant, then we will have direct proportion between density and pressure        Day 16 ­ 2/10/2016  Phys 5b    Waves  D(x, t) = D​m​in(kx ­ wt)  x ← D = x​ ­part​ o  D = 0 when particle pass through x​ o​for a given particle    How does P oscillate?    Speed of the Wave  Elastic prop vw​= √ Inertial prop ← ρ  String vw​=  Ts  √ μ B = − ΔP ΔP  =− B ΔV   ΔV /V V given ΔP  →  ΔV ↑ → B↓  V speed of sound c​ s​= √ ρ  gas B↓  5 kg air  Bair​ 1.5 x 10​ cs​= 343 m/s ρ  =  1.2m 3  9​ Ex. B​w​= 2.2 x 10​ Pa  ← c​w​= 1482 m/s    Take a volume of the gas V  ΔV AsΔD V = A​s​Δx ΔV = A​ sΔD ΔP  = − B V  = − B AsΔx  P​max​D​ m​k  c​ =  B   s​ √ ρ B = c​s​ w 2πf k = cs=  cs  ΔP max = 2πc sD m   Ex.   c​ = 340 m/s  s​ 3 ρ = 1.28 kg/m​   ­1 f = 2000s​   D​  = 100nm  m​ ΔP​ max​55 Pa → ( N2)  m Tube  cs fn​ n 4L n = 1, 3, 5, …  f =  cs   1​ 41 f = f​ = cs   3​ 1​ 4L1 Gas Behavior  1. P​ o​ρo vo​= 0  2. No gravity  3. small displacements & only in x­direction  4. only small variations in time & space of perturbations  ρ  =  ρo+ ρ o v = v​1  dP1 dv1 dt + ρo dx = 0  dv dv 1dP Momentum Equation: dt=   − v dx− ρ dx    Day 17 ­ 2/12/2016  Phys 5B    Linearization Process    dρ1 dv1 dt +  ρo dx = 0  dv1 dP1 ρ odx  =  ­  dt   dv1 dP1dρ0 dP 1 dP1dρ1 ρ o dt =− dρ dx dx = dρ dx   1 1 2 2 dρ1 dv1 d d o d v1 dt =  − ρ odx  ← take derivative   →  dt dt2 =   − ρodxdt  dP dρ dv dP d ρ d v − 1 0  =  ρ 1  ← take derivative  d  → − 1 2 = ρ o 1  dρ1dx o dt dx dρ1dx dxdt d ρo dP 1 ρ 1 2 1 d ρ1 add the two derivatives →  2  ­  dρ dx2  → d ρ  =o dP  / dρ 2   dt 1 1 1 dt units: [   ] =  N 1  = kg m m  = m​2 / s​  ρ m kg / m 3 s kg d ρ 1 d ρ1 dx  =  s dt     Sound Speed  2​ dP1 dP1 cs​ =  dρ1 c2​=  √ dρ1   PV = NkT k = 1.38 x 10​  ­23 P = S mk T  p dP1 k k dρ1 =  mp T → c​ s​=  mpT    √   String  T​  = T​ α → α = tanα  sy​ s T​=  ­ T​   dy sy ​ sdx P(t, x) = A​ 2T​kwcos​ 2(kx ­ wt)  s​ P  =  T​ A​2kw  2 s​ w w vw​ =  k→ k =  vw   vw​ =  T s → T​ s​= v w  √ μ 1 2 2 w 1   2 2 P =   v2μw vww =  v 2Aww   2 2 2 P  = 2π  v ρw A A   s   Wave Intensity  P 2 2 2 I  =  As 2π v ρfwA   2 P A​s​= 4πr  → I =  4πr    Inverse square law r goes up, I goes down    Interference  y 1 =  Asin(kx 1 wt  +  ϕ )01 y 2 =  Asin(kx 2 wt  +  ϕ )02 Δϕ  =  ϕ  2− ϕ  1 k(x   −2 x ) 1+  ϕ  o2 ϕ   01 Δρ  =  kΔx  +  Δϕ  →02hase length, path length difference  2π Δϕ  =   Δλ + Δϕ   o if:  Δϕ  =  2nπ perfect constructive  Δϕ  =  (2n + 1)π  perfect destructive interference

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