1Lecture 26Reminders:Reading for this week TZD through 7Homework #9 due next WednesdayExam #2 on Friday, October 30 in class.Details on class web pageToday’s topic:Wave Equationsand Solutionshttp://tinyurl.com/2170WeeklySurveyAs an additional assignment for Homework #9, please complete this survey related to the course. It only takes ~ 15 minutes and provides valuable information. It is worth one homework problem, but your answers are anonymous.Please fill this out by Wednesday next week (October 28).This link is now included on the class web page and homework page.Vibrations on a string: Electromagnetic waves: Example Wave Equations You Have Seen (or not)222221tyvxy∂∂=∂∂v=speed of wave222221tEcxE∂∂=∂∂c = speed of lightxyExMagnitude is non-spatial:= Strength of Electric fieldMagnitude is spatial:= Vertical displacement of StringSolutions: E(x,t) Solutions: y(x,t)Solving the Standard Wave Equation1. Guess the function form(s) of the solution2. Plug into differential equation to check for correctness, find any constraints on constants3. Need as many independent functions as there are derivatives.4. Apply all boundary conditions (more constraints on constants)222221tyvxy∂∂=∂∂Physics laws are often differential equations.F=ma can be written as22txmF∂∂=kxtxm −=∂∂22If the force is a restoring force so F = −kx thenThe standard wave equation isGeneric prescription for solving differential equations in physics:A.B.C.D.E. More than one of the aboveClicker Question222221tyvxy∂∂=∂∂The standard wave equation is Step 1: Guessing the functional form of the solutionWhich of the following function forms is a possible solution to this differential equation?22tAxy =)cos(BxAy =)cos(BtAy =)sin()cos( CtBxAy =The A. form leads to or222212 AxvAt =222vxt =Different functions on the left side and right side. This is incorrect.The B. form leads to which doesn’t work.0)cos(2=− BxABThe C. form leads to which doesn’t work.)cos(02BtAB−=Claim that is a solution toStep 2: Check Solution and Find Constraints)sin()cos( CtBxAy=222221tyvxy∂∂=∂∂Time to check the solution and see what constraints we haveLHS:)sin()cos(222CtBxABxy−=∂∂RHS:)sin()cos(122222CtBxvACtyv−=∂∂Setting LHS = RHS:)sin()cos()sin()cos(222CtBxvACCtBxAB −=−This works as long as222vCB =We normally write this as)sin()cos( tkxAyω=so this constraint just means222vkω=orfkvλω==2and we have the constraint thatSince the wave equation has two derivatives, there must be two independent functional forms.Constructing General Solution from Independent Functions222221tyvxy∂∂=∂∂)sin()cos( tkxAyω=222vkω=)cos()sin( tkxByω=The general solution is)sin()cos()cos()sin( tkxBtkxAyωω+=Can also be written as)sin()sin( tkxDtkxCyωω++−=xyt=0We have finished steps 1, 2, & 3 of solving the differential equation.Last step is applying boundary conditions. This is the part that actually depends on the details of the problem.* traveling waves in opposite directionsBoundary Conditions for Guitar String0LGuitar string is fixed at x=0 and x=L.222221tyvxy∂∂=∂∂Wave equationFunctional form:)sin()cos()cos()sin( tkxBtkxAyωω+=Boundary conditions are that y(x,t)=0 at x=0 and x=L.Requiring y=0 when x=0 means)sin()0cos()cos()0sin(0 tBtAωω+=which is)sin(00 tBω+=This only works if B=0.So this means)cos()sin( tkxAyω=Clicker questionBoundary conditions require y(x,t)=0 at x=0 & x=L. We found for y(x,t)=0 at x=0 we need B=0 so our solution is . By evaluating y(x,t) at x=L, derive the possible values for k.A. k can have any value B. π/(2L), π/L, 3π/(2L), 2π/L …C. π/LD. π/L, 2π/L, 3π/L, 4π/L …E. 2L, 2L/2, 2L/3, 2L/4, …. )cos()sin( tkxAyω=To have y(x,t) = 0 at x = L we need0)cos()sin( =tkLAωThis means that we need 0)sin( =kLThis is true for kL = nπ. That is,Lnkπ=n=1n=2n=3So the boundary conditions quantize k.This also quantizes ωbecause of the other constraint we have:222vkω=Summary of our Wave Equation Solution1. Found the general solution to the wave equation222221tyvxy∂∂=∂∂)cos()sin()sin()cos( tkxBtkxAyωω+=)sin()sin( tkxDtkxCyωω++−=or2. Put solution into wave equation to get constraint222vkω=3. Have two independent functional forms for two derivatives4. Applied boundary conditions for guitar string. y(x,t) = 0 at x=0 and x=L. Found that B=0 and k=nπ/L.Our final result:)sin()cos( tkxAyω=222vkω=Lnkπ=with andn=1n=2n=3Standing wavesStanding waveStanding wave constructed from two traveling waves moving in opposite