Homework 11 collaboration area
When is this homework due? I don't see any annoucement on the webpage.
From Eun Young:
it's due Wed. 11/20 (See Lesson 36).
From Farhan: Any hint on how to go about #16 of 12.3?
From Eun Young :
We already have F(x) and G(t) from #15.
Using the given conditions, we need to compute coefficients and find $ \beta $.
We have $ F(0)=F(L)=F^{''}(0)= F(L)^{''}= 0. $
Using $ F(0)=F^{''}(0)=0 $, we can show that the coefficients of cos and cosh functions are zero.
Using $ F(L)=F^{''}(L)=0 $, we can show that the coefficient of sinh is zero.
Hence, $ F(X) = \sin (\beta x) $.
Using $ F^{''}(L)=0 $, we can find $ \beta $.
Plug this $ \beta $ into G(t) and use the zero initial velocity condition, then we'll get G(t).
From Craig:
For #15 on 12.3, are we supposed to show the work for each of the end conditions, or only part a (simply supported)?
From Eun Young:
You do not need boundary conditions for #15. See Lesson 38 to get some hints.
Question by Ryan Russon: For #8 of p. 556, I am having difficulties finding the solution for this in terms of what should happen with t... I realize that it must meet the IC's and the BC's but I can't figure out a periodic type solution that would vibrate for t>0 Thanks!
For #16, I am just not sure how u can eliminate the 3 constants in front of cos, cosh, and sinh to be zero... Just to be sure, we equate F(0) and F(0) and equate F(L) and F(L) correct? With this I get that A+C = -(beta^2)*(A+C). How can we assume both A and C are zero? For the second equation, I am not at all sure how the sinh coefficient goes to zero...I am getting A = -A*beta^2, B = -B*beta^2, and so forth...suggestions?
--From Collier Miers: Just solve F(0)=0 to get the value of C. Then plug that C into the equation for the second derivative of F at x=0. This will allow you to determine that A and C are equal to zero. Then do essentially the same process for F(L) and its corresponding second derivative. Hope this helps.
Question by Ryan Leemhuis: For #19 of p. 542, I get the characteristic equation theta+y^2 = 0 with a double root at -y^2. However, it appears that the solution should be the integral of this based on the y^3/3 in the back of the book. Any advice?
From Steve Bell: Ryan, I showed in class one day how you could solve that kind of problem like a first order linear ODE with a parameter (the other variable) floating along. Another way to do it is to notice that it is a separable ODE. Bring all the u's and du's to one side and all the y's and dy's to the other and integrate. The arbitrary constant should be taken to be an arbitrary FUNCTION of the floating along variable (x in this case).
--From Hzillmer Any tips for number 11 of 12.3? I keep getting 0 for u(x,t), clearly wrong. Thanks
--From Dori: Hzillmer, I used the formula the Professor wrote out in the second part of Lesson 36 for u(x,t) with the An and Bn coefficients. Bn should be zero, but for An I took the integral from 0 to 1 and split it into 4 integrals (0 to 1/4, 1/4 to 1/2, etc.). The first and last integral are zero but you can get an f(x) for the middle two from the graph. It's sort of a bear to integrate but the cosine terms end up cancelling out and you're left with all sine terms. Then you plug that into the u(x,t) formula. At least that's what I did, someone correct me if I'm wrong.
--Question on 12.3 #15 - I'm confused on why -G/c^2*G = a constant and why it's shown as B''? Also, I don't understand how cosh and sinh show up for F(x). Any hints? Thanks, Tlouvar
From Steve Bell: I talked about this at the beginning of lecture on Friday. When you separate variables, you'll get all the functions of x on one side and all the functions of t on the other. In order for a function of x to be equal to a function of t, they both must be equal to the same constant function. The fourth power of beta just makes the solution to the 4th order ODE easier to write out.
