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By Robert Hermann

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When the string is plucked we can approximately describe its vertical displacement from equilibrium at each point x as a function y(x, t). 11). We will assume no external forces and that horizontal acceleration is negligible. It follows from Newton’s second law that • difference in vertical tensions: (T sin θ)| x+Δx − (T sin θ)| x = Δm ytt , • difference in horizontal tensions: (T cos θ)| x+Δx − (T cos θ)| x = 0. 10 Vibrations of a long string. 11 Tension forces acting on a small segment of the string.

9 Introduction Hint: Set ξ = x − ct, integrate twice with respect to ξ, and use uξ = q(u) to solve the resulting equation. Consider the sine–Gordon equation u xx − utt = sin u. (a) Using the transformation χ = γ(x − vt), τ = γ(t − vx) write the equation in terms of the new coordinates χ, τ; find γ in terms of v, −1 < v < 1 so that the equation is invariant under the transformation. (b) Consider the transformation ξ = (x + t)/2, η = (x − t)/2. Find the equation in terms of the new coordinates ξ, η.

8) 48 Asymptotic analysis of wave equations √ where A(x/t)/ t is the decaying in time, slowly varying, complex amplitude. √ This decay rate, O(1/ t), is slow, especially when compared to those cases when the solution decays exponentially or as O(1/tn ) for n large. Note that dimensionally, ω corresponds to a speed. Stationary phase has shown that the leading-order contribution comes from a region moving with speed ω , which is termed the group velocity. This is the velocity of a slowly varying packet of waves.

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