Consider the vibration of an infinite string: Utt = 4uxx, - < 0 u(x,0) = f(x)...

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Consider the vibration of an infinite string: Utt = 4uxx, -∞<<∞, t> 0 u(x,0) = f(x) ut(x, 0) = g(x) Sketch the solution at t = 1 and t = 2 using the x-t plane display similar to those from the lecture notes. (a) f(x) = H (x + 1) - H(x - 1), g(x) = 0. (b) f(x) = H(x), g(x) = 0. (c) f(x) = 0, g(x) = H (x + 1) - H(x - 1). (d) f(x) = 0, g(x) = H (x). where H(r) is the Heaviside step function H(x) = Jo, x<0 1, x>0 Consider the vibration of an infinite string: Utt = 4uxx, -∞<<∞, t> 0 u(x,0) = f(x) ut(x, 0) = g(x) Sketch the solution at t = 1 and t = 2 using the x-t plane display similar to those from the lecture notes. (a) f(x) = H (x + 1) - H(x - 1), g(x) = 0. (b) f(x) = H(x), g(x) = 0. (c) f(x) = 0, g(x) = H (x + 1) - H(x - 1). (d) f(x) = 0, g(x) = H (x). where H(r) is the Heaviside step function H(x) = Jo, x<0 1, x>0

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a Given fx Hx1 Hx1 gx 0 The dAlembert solution is uxt 12fxt fxt At t1 ux1 12fx1 ... View the full answer