The string is initially at rest, so . Taking the Laplace transform in converts the wave equation intoFor , the solution that decays as is . The transformed boundary condition fixes , and therefore
For ,The Bromwich inversion formula givesWhen , close the contour to the left. The residues at sum to . When , close it to the right, where there are no poles, and the integral vanishes. HenceEquivalently, in terms of the Heaviside step function. The boundary oscillation travels to the right at wave speed without reflection or dispersion; a point at begins moving after the propagation delay .
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