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Unformatted text preview: PHYS 218  SOLUTION TO ASSIGNMENT 4 23 Feb 2007 By Chung Koo Kim email : [email protected] 1. String with a massive bead Let’s suppose that the sinusoidal wave is incoming from the left. The wave equations of left and right side of the mass can be written η left = e ikx + Re ikx η right = Te ikx so that the η left has both incoming/reflected components and η right has only outgoing transmitted component. ♣ Remarks. The following forms may be more general and familiar: η left = A 1 e i ( kx ωt ) + B 1 e i ( kx + ωt ) η right = A 2 e i ( kx ωt ) Pay attention to the signs of kx and ωt . Waves propagating toward the positive x direction should have opposite sign for kx and ωt , and same sign for negative x direction. And we may replace the i in the above equation by i without loss of generality at least for our purpose. But for later consistency’s sake, particularly in quantum mechanics, we keep + ikx for wave toward (+x) and ikx for (x). You will learn that e ikx has positive momentum when we apply momentum operator ˆ p = ¯ h i ∂ ∂x . And since we are mostly interested in the relative ratio of the coefficients only, we divide the equation by A 1 and replace B 1 /A 1 = R and A 2 /A 1 = T . Of course we took the common factor of e iωt out. Then we get the first set of equations. ♣ To match the two solutions at x=0, we should find two boundary conditions: η left ( x = 0) = η right ( x = 0) τ ∂η left ∂x x =0 + τ ∂η right ∂x x =0 = m ∂ 2 η right ∂t 2 The first one comes from the continuity of the string, and the second one from Newton’s second law applied to the vertical component of force. The η ( x, t ) in the second equation can be either η left ( x, t ) or η right ( x, t ) since the two are the same at x=0. Note that we are in practice thinking of the limit approaching zero from the left and right, when plugging x=0....
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 Spring '08
 WITTICH,P
 Energy, Mass, k2, Bulk Modulus

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