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by: Shanee Dinay

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# Week 4 Notes PHYS 5B

Shanee Dinay
UCSC
GPA 3.94

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Week 4 of notes for Intro to Physics 2. Topics include Oscillators, Simple Pendulums, and Damping.
COURSE
Intro to Physics II
PROF.
A.Steinacker
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
physics 2, Harmonic Oscillators, Damping, Simple Pendulum
KARMA
25 ?

## Popular in Physics 2

This 6 page Class Notes was uploaded by Shanee Dinay on Monday February 8, 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 12 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.

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Date Created: 02/08/16
Day 9 ­ 1/25/2016    Homework ­ sketch, solutions are available online  Oscillators  d x 2 Equation dt2+ w x = 0  d x k k dt2+ xm= 0 w =  √ m      x(t) = Rcosθ x =  dθ → θ = wt  dt x(t) = Rcos(wt) w = omega  v(t) = ­Rsin(wt)  2​ a(t) = ­Rw​ cos(wt)  2π = ωT Period T =  2π θ = wt  ω       Spring  Fe​= F​G ​t part 2  ma = F​ e​ F​G   2 d 2  = ­ky + mg  dt d t k d y k dt +  m ­ g = 0 but  dt + m = 0  F​eo​ F​G​= 0  k ­ky​0​ mg = 0 → g =  y   m o 2 2 d 2  +  y ­  y  = 0 →  d (y2−oy +  (y − y )  =  0 y ­ y​ = u  dt m m o dt m o 0​ d u k k →  dt2+ um= 0 ω = √ m    E​K​+ E​PE​constant  mv 2 kx2 (t+ (t = constant  2 2 2 2 2mv(t) dv( +  2kx(t)dx  = 0 → mv(t) d x  + kx(t) = 0 →  d 2  +  x = 0  2 dt 2 dt v(t) dt dt m x(t) = xo​os(wt)  v(t) = ­x0​sin(wt)  mv2 mx ow sin (ωt) kx k 2 2 E​k​t) =  2  =  2 E​PE​ =  2  = 2x oos (ωt)  The Simple Pendulum    F​G​ ­mgsinθ    positive torque τ = Iα  τ = r x F = r mgsinθ  d θ 2 = ­ Lmgsinθ =  dt2mL   d θ g   dt  + Lsinθ = 0   For small angles θ  d θ g g L sinθ ~ θ radians   dt  + Lθ = 0 ω =  L T = π √ g   The Physical Pendulum  next class    Day 10 ­ 1/27/2016  PHYS 5B    Harmonic Oscillators  2 d 2  + ω x = 0 ω   =     2t m d (y −o  +  (− y ) = 0  dt m o d θ g dt  +  L = 0  simple pendulum  d θ mgr dt  +  I θ physical pendulum  Ex. Rod    τ < 0 ω o mgr    √ I I​ =  mL 2 Io= I + mr   CM​ 12 C2 2 2 r =  − 1     = mL  + mL  =  mL   2 3 12 36 9 mg(L/6) 2g 2π 2L ω o=  √ m(L /9) = √ 3L   T =  ω o = 2π √ 3g   Ex    Mgr 2 2 2 ω o I ICM   =  2 Io  =  m2 + MR   o 2 2 2      =mR + mR = 2mR   2 4 4 mg(R/2) 2 g 2π 3R ω o √ (3/4)mR  = √ 3R   TB​=  ωo = 2π √ 2g   2 d 2  + ω x = 0   dt x’’ + P​x’ + q​ y = 0  o​ o​ at Po​= Ansatz → x(t) = e​   2 αt αt αt 2 α e + P αeo+ qe = 0 characteristic equation α + P αo+ q = 0   α t α t x(t) = Ae   +  Be 2 A​1​ determined from initial condition   2 −Po±√ Po−yq a1, 2​ 2   Ex. z = x + iy    |z| =  x + y   tanθ  =     y √ x z* = x ­ iy conjugate  z•z* = (x + iy)(x ­ iy) = x​ + y​ = z​2  Exponential form of a Complex Number  ∞ 2n+1 ∞ 2n sinx ≃   ∑ (ix) cosx ≃   ∑ (ix)  i (2n+1)! (2n)! n=0 ∞ n=0 ix n cos x + isin x =  ∑( )n! n=0 ix​ ∞ xn iθ e​  =  ∑ n! Euler formula: e   =  cosθ  +  isinθ  =  z  n=0 iθ e   =  cosθ  −  isinθ  =  z*  iωot −iωot x(t) = Ae + Be   x(t) must be real x(t) = x*(t)  Ae iωot+ Be−iωot = A e −io + B e iωo  * * x(t) = 2acos(w​ t) ­ 2bsin(w​ t)  o​ o​ x(0) = x​  = 2a a =  xo amplitude of displacement  o​ 2 x(t) = x​cos(w​ t)  o​ o​ Damping  d x m dt2 = ­kx + F​D FD​ = ­bv  d x dx m dt  + b dt + kx = 0  2 kgm d 2 +  b dx  +  x = 0 b damping const [b] =  NS  =  2 •  s   = frequency  dt m dt m m s m m characteristic equation  α + α + ω = o  m now find roots to solve  a​ =  − b ± ( ) − ω   2  1, 2​ 2m √ 2m o ( b )   <  ω   2m o   Day 11 ­ 1/29/2015  Phys 5b    Damping  d x b dx k dt+ m dt+ m  = 0  F​D​= ­bv Drag Force  2​ k w​o​ = m  B =  b   2m x’’ + 2Bx’ + w​ ’x = 0 → x = e​at  o​ α​ + 2Bα + w​ o​ = 0  2 →  α = 2B ±4B −4o  1, 2 √ 2 1. case 1 ­Δ < 0  ← → B < w​   b < 2mw​   o o 2 2 2 β − ω  o = − ωo− β    √ √ α1, 2​B ±iw​ o α1t​ α2t  x(t) = Ae​ + Be​     ​ (­B +iw)​ (­B­iw)   ​ Ae​ t + Be​   x(t=0) = x​ A+B = x​   o o v(t=0) = 0 v = x’ = A(­B + iw​1​e​. + B(­B ­ iw1​e​. t v(0) = 0 = ­A​B​ + iw1​ ­ B​  iw​     = ­B(A+B) + iw​ (a­B)   1​ → A ­ B =  βxo  w 0 x(t) = e​(Ae​ + Be​) iwt​ ­Bt​ β → X(t) = x​o​​[cost(w​ 1​) + w sin(w​1​)] ← general solution for harmonic oscillator  1 β = b ω   =   k   2 2  2m 0 √ m w 1 √ w o β   v kx   E = E​K​+ E​P​= m 2+  2   dE m dv k dx dv dt=  2v  dt 2x2 dt = mv  dtkxv  dE = dW​ FD  dW​ = F​ •dx = bv • dx  FD​dW D​ dE =  FD   F​D​dx  = FD​• v (Power)  dt dt dt Drag force  dE 2 dt =   − bv   E Quality Factor: Q = 2π ΔE   b 2. case 2 β > 2mw o β = 2m   Strong Damping  2 2 2 1 a1, 2​ − β ± √ β − ω  0 w1​= √ β − ω o  (­B + w1)t​ −β−w )t x(t) = Ae​  + βe 1   A =  (1  +   )β B =   (1  −   )   2 w1 2 w1 x0 −βt β w1t​ β ­w1t x(t) =  2 e [(1 +  w1 )e​ + (1­ w1 )e​    3. case 3  Δ  =  0 β = ω b = ω   0 2m o −βt x(t) = x​0​1 + βt)e

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