ph2a_quiz2_soln

Since the string has no initial velocity the phases

  • Notes
  • AgentStarOkapi9220
  • 4

This preview shows 1 out of 3 pages.

1, 2, 3, 4 at the end (this ultimately will require less work). Since the string has no initial velocity, the phases δ n can be chosen to be zero and the coefficients are given by A n = 2 L L 0 y ( x , t = 0 ) sin n π x L dx . 1
Image of page 1

Subscribe to view the full document.

Because we are dealing with a piecewise-continuous function the integral breaks into the following parts A n = 2 L L /4 0 4 h L x sin n π x L dx + 2 L L /2 L /4 2 h sin n π x L dx - 2 L L /2 L /4 4 h L x sin n π x L dx . Let’s evaluate each integral separately, and add the results to get the final answer. Inte- grating by parts, one has that 2 L L /4 0 4 h L x sin n π x L dx = - 8 h n π L x cos n π x L x = L /4 x = 0 - L /4 0 cos n π x L . One can then evaluate this integral to find 2 L L /4 0 4 h L x sin n π x L dx = - 2 h n π cos n π 4 - 4 n π sin n π 4 . The second integral can be evaluated immediately to find 2 L L /2 L /4 2 h sin n π x L dx = - 4 h n π cos n π 2 - cos n π 4 . The third integral is similar to the first, and by integrating by parts, one can find that - 2 L L /2 L /4 4 h L x sin n π x L dx = 4 h n π cos n π 2 - 1 2 cos n π 4 - 8 h n 2 π 2 sin n π 2 - sin n π 4 . There is a large amount of cancellation when the three integrals are added, leaving just A n = 8 h n 2 π 2 2 sin n π 4 - sin n π
Image of page 2
Image of page 3
You've reached the end of this preview.

{[ 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