ps9_soln - Physics 105 Problem Set 9 Solutions Problem 1 (3...

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Physics 105 Problem Set 9 Solutions Problem 1 (3 pts) Motion of the support launches a wave traveling along the string with the speed v = s T s μ . (a) An excitation of the support at x = 0 reaches a point with coordinate x with time delay Δ t = x/v . Therefore the displacement is D ( x,t ) = A sin π t - x/v τ · , 0 t - x/v τ ; 0 otherwise . (b) All the kinetic energy is contained in the moving part of the string: K = 1 2 μ Z v ( t + τ ) vt dx ± ∂D ∂t 2 . Substituting for D ( x,t ) and making a shift of the integration variable yields K = μ π 2 A 2 2 τ 2 Z 0 cos 2 πx dx = μπ 2 vA 2 4 τ = π 2 A 2 μT s 4 τ . (c) As it was derived in the precept, potential energy of the string is U = T s Z s 1 + ± ∂D ∂x 2 - 1 · dx 1 2 T s Z ± ∂D ∂x 2 dx ; where we have used the fact that the displacement is small. Evluating this integral we get U = π 2 A 2 T 4 = π 2 A 2 μT s 4 τ . Note that the amounts of kinetic and potential energy contained in the string are identical. (d) Let us stress that we cannot simply say that the amplitude A has to be “small” since in is not a unitless quantity (unlike ∂D/∂x
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This note was uploaded on 11/05/2011 for the course PHY 105 at Princeton.

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ps9_soln - Physics 105 Problem Set 9 Solutions Problem 1 (3...

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