Question

A string, such as described in the problem above, can be initially distorted into an arbitrary shape, subject to the B.C. applied at x-0, L. If you assume the shape is a triangle, can you find the components in terms of the sine functions? A) Write an expression for each component. B) See if you can compute the coefficients of each sine function. (This is a Fourier sine series expansion of the "triangle" function.)

Please help me solve question 4

Answer 1

Let us assume it is an equilateral triangle such that at x = 0 & L y = 0 f(x) = y = {squareroot 3 x 0 lessthanorequalto x lessthanorequalto L/2 -squareroot 3 x + squareroot 3 L L/2 lessthanorequalto x lessthanorequalto L f(x) = a_0/2 + sigma^infinity_n = 1 a_n cos(2 n pi x/L) + sigma^infinity_n = 1 b_n sin(2 n pi x/L) a_0 = 2/L integral^L_0 f(x)dx = 2/L[integral^L/2_0 squareroot 3 x dx + integral^L_L/2(squareroot 3 L - squareroot 3 x)dx] a_0 = 2 squareroot 3/L[L^2/8 + L^2/2 - 1/2(3L^2/4)] = squareroot 3/2 L a_n = 2/L integral^L_0 f(x) cos(2 n pi x/L)dx a_n = 2 squareroot 3/L(integral^L/2_0 x cos(2 n pi x/L)dx + integral^L_L/2(L - x) cos(2 n pi x/L)dx} a_n = 2 squareroot 3/L{L^2/4 n^2 pi^2(cos n pi - 1) - L^2/4 n^2 pi^2(cos 2 n pi - cos n pi)} a_n = 2 squareroot 3 L/4 n^2 pi^2{2 ((-1)^n - 1)} \r\n a_n {-2 squareroot 3L/n^2 pi^2 n = odd 0
i got cosine series may be i choose wrong triangle.please provide the figure to correct it.