5soln5 - Physics 315: Oscillations and Waves Homework 5:...

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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 dx . Furthermore, the elements transverse velocity is y ( x, t ) /t . Hence, the kinetic energy of the element is dK = 1 2 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 parenleftbigg y t parenrightbigg 2 dx. (2) The potential energy of the string when its extension is 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 s is U = integraldisplay s T ds = T 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 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, 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 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 x l parenrightBig cos parenleftbigg n v t l n parenrightbigg . (9) It follows from (2) that K...
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5soln5 - Physics 315: Oscillations and Waves Homework 5:...

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