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Unformatted text preview: Solutions for 2009 Midterm Exam Phys. 12a H. J. Kimble (Dated: 3 November 2009) PROBLEM 1: EIGENFREQUENCIES The general solution to the one dimensional wave equation is (Textbook Chapter 2 Eq.(23)) ψ ( z,t ) = cos( ωt + φ )[ A sin( kz ) + B cos( kz )] . (1) We need to determine the wavenumber k = 2 π λ for various boundary conditions. (We only look for the nontrivial solutions for which k negationslash = 0.) Let’s call f ( z ) = A sin( kz ) + B cos( kz ) . (a). We have f ( z ) = 0, for z = 0 ,L . Therefore, we have B = 0 and sin( k p L ) = 0. Hence k p = pπ L , with p = 1 , 2 , 3 , ··· . (b). We have df/dz = 0, for z = 0 ,L . Therefore, we have A = 0 and sin( k p L ) = 0. Hence the solutions for k p are the same as in (a), namely k p = pπ L , with p = 1 , 2 , 3 , ··· . (c). This time, we need f ( z ) = 0 at z = 0 and df/dz = 0 at z = L . So B = 0 and k p = ( p + 1 2 ) π L , with p = 0 , 1 , 2 , ··· . (d). Finally, we require f ( z + L ) = f ( z ). The only way to achieve this is requiring k p L = 2 pπ . So k p = 2 pπ L , with p = 1 , 2 , 3 , ··· . For all cases, by way of the dispersion relation, we have ω p = v k p = radicalBig T ρ k p PROBLEM 2: AN ELECTRICAL FILTER Consider an electrical filter consisting of a series of coupled capacitors and inductors, a section of which is shown below. L C C C I n 1 I n I n+ 1 L L a V in V out (a) Show that the equations of motion are d 2 dt 2 ( I n 1 + I n +1 − 2 I n ) = I n LC . (2) 2 The current flowing down through the n th inductor is I n 1 − I n , so that the voltage across it is V L,n = − L d dt ( I n 1 − I n ) . (3) The voltage across the the n th capacitor is V C,n = Q n /C so that dV C,n dt = I n C . (4) But V C,n = V L,n +1 − V L,n . Differentiating Eq. (3) and equating voltages, we obtain Eq. (2). (b) By trying a solution of the form I n = A cos ωt sin nka, (5) or otherwise, find the dispersion relation of the system in the dispersive band....
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 Spring '09
 DUDKO
 mechanics, Equations, Frequency, Trigraph

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