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Unformatted text preview: Physics 315: Oscillations and Waves Homework 5: Solutions 1. Consider a small element of the string lying between x and x + dx . The mass of the element is ρ dx . Furthermore, the element’s transverse velocity is ∂y ( x, t ) /∂t . Hence, the kinetic energy of the element is dK = 1 2 ρ dx bracketleftbigg ∂y ( x, t ) ∂t bracketrightbigg 2 . (1) Summing over all elements, the total kinetic energy of the string becomes K = integraldisplay l dK = 1 2 integraldisplay l ρ parenleftbigg ∂y ∂t parenrightbigg 2 dx. (2) The potential energy of the string when its extension is Δ s is equal to the net work done in stretching the string. Now, when the string is in tension, it pulls back on its supports with a constant force T . Hence, the work done in increasing the extension by ds is T ds . It follows that the net work done in producing an extension Δ s is U = integraldisplay Δ s T ds = T Δ s. (3) Now, an element of length of the string is written dl = ( dx 2 + dy 2 ) 1 / 2 , or dl = bracketleftBigg 1 + parenleftbigg ∂y ∂x parenrightbigg 2 bracketrightBigg 1 / 2 dx. (4) Thus, the net extension of the string ( i.e. , the difference between its stretched and unstretched lengths) is Δ s = integraldisplay l dl − l = integraldisplay l bracketleftBigg 1 + parenleftbigg ∂y ∂x parenrightbigg 2 bracketrightBigg 1 / 2 − 1 dx. (5) However, for small amplitude transverse displacements, we can treat ∂y/∂x as a small quantity. Thus, making use of a binomial expansion, Δ s ≃ integraldisplay l parenleftBiggbracketleftBigg 1 + 1 2 parenleftbigg ∂y ∂x parenrightbigg 2 + ··· bracketrightBigg − 1 parenrightBigg dx ≃ 1 2 integraldisplay l parenleftbigg ∂y ∂x parenrightbigg 2 dx, (6) and so U = 1 2 integraldisplay l T parenleftbigg ∂y ∂x parenrightbigg 2 dx. (7) Finally, the total energy of the string is E = K + U = 1 2 integraldisplay l bracketleftBigg ρ parenleftbigg ∂y ∂t parenrightbigg 2 + T parenleftbigg ∂y ∂x parenrightbigg 2 bracketrightBigg dx. (8) Now, the general motion of the string can be represented as a linear super position of normal modes: i.e. , y ( x, t ) = summationdisplay n =1 , ∞ A n sin parenleftBig n π x l parenrightBig cos parenleftbigg n π v t l − φ n parenrightbigg . (9) It follows from (2) that K...
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 Spring '10
 Fitzpatrick
 Energy, Force, Kinetic Energy, Mass, Work, Trigraph, Boundary value problem, Normal mode

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