ph2a_quiz2_soln

2 this allows one to see that a 1 8 h π 2 2 1 a 2 4

Info icon This preview shows pages 2–4. Sign up to view the full content.

View Full Document Right Arrow Icon
2 . This allows one to see that A 1 = 8 h π 2 ( 2 - 1 ) , A 2 = 4 h π 2 , A 3 = 8 h 9 π 2 ( 2 + 1 ) , A 4 = 0. Part 4 [2 points] : Are any of these first four coefficients zero? If so, can you think of a reason why? (Hint: Think about the symmetry of the characteristic modes.) The coefficient A 4 is zero, for the following reason. Because the initial displacement is zero from L /2 x L , only the region between 0 and L /2 will contribute to the Fourier coefficients. Because the initial displacement of the string is an even function about x = L /4 while the 4 th mode sin ( 4 π x / L ) is odd about x = L /4, their product is an odd function, and this coefficient must vanish. (As an aside, one can see that this holds true for all modes of the form A 4 n , n N either by inspecting the formula of part 3, or through an argument of the type given above). 2
Image of page 2

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Part 5 [2 points] : Sketch the approximate for y ( x , t = 0 ) obtained by using only the first four terms in the series expansion. At time t = 0, the Fourier series is given by y ( x , t = 0 ) = n = 1 A n sin n π x L , for the coefficients A n found in part 3. Since the series is being truncated after the first four terms, one has that y ( x , t = 0 ) 8 h π 2 ( 2 - 1 ) sin π x L + 4 h π 2 sin 2 π x L + 8 h 9 π 2 ( 2 + 1 ) sin
Image of page 3
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern