Question

Torsional vibration of a shaft is governed by the wave equation,

Torsional vibration of a shaft is governed by the wave equation, 9 where 0(x, t) is the angular displacement (angle of twist) along the shaft, x is the distance from the end of the shaft and t is time. For a shaft of length 2T that is supported by frictionless bearings at each end, the boundary conditions are 0 x(0,t) = 0x(2T, t) = 0, t> 0 Suppose that the initial angular displacement and angular velocity are O(x,0) = 6 cos(4x), 0t(x,0) = 5+3 cos(4x), 0 x2T, respectively. You may use the result that the eigenvalues of the boundary-value problem x'(0) = X'(27) = 0 X" -X 0, are n2 = 0,1,2,3, ... n 22 with corresponding eigenfunctions

Answer 1

"B. C: Given: partial differential^2 theta/partial differential t^2 = g partial differential^2 theta/partial differential x^2: theta_x(0, t) = theta_x(2pi, t) = 0, t > 0 I. C: theta(x, 0) = 6 cos 4x theta_t(x, 0) = 5 = 3 cos 4x. bullet Let theta(x, t) = X(x). T(t) partial differential^2 theta/partial differential t^2 = X T"": where T"" = partial differential^2 T/partial differential t^2 parialdifferential^2 theta/partial differential x^2 = X"" T: where X"" = partial differential^2 X/partial differential x^2. Plugging the values above in P. D. E we get XT"" = g X"" T Dividing by XT throughout, we have: X""/X = 1/g T""/T = -lambda (say) solving X""/X = -lambda we have X(x) = c_1 cos squareroot lambda x + c_2 sin squareroot lambda x solving 1/g T""/T = -lambda we have, T(t) = c_3 cos 3 squareroot lambda t + c_4 sin 3 squareroot lambda t Now theta(x, t) = X(x). T(t) theta(x, t) = (c_1 cos squareroot lambda x + c_2 "

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