p33_028 - 28(a In Eq 33-25 we set q = 0 and t = 0 to obtain...

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28. (a) In Eq. 33-25, we set q =0and t = 0 to obtain 0 = Q cos φ .T h i sg i v e s φ = ± π/ 2 (assuming Q 6 = 0). It should be noted that other roots are possible (for instance, cos(3 π/ 2) = 0) but the ± π/ 2 choices for the phase constant are in some sense the “simplest.” We choose φ = π/ 2tomakethe manipulation of signs in the expressions below easier to follow. To simplify the work in part (b), we note that cos( ω 0 t π/ 2) = sin( ω 0 t ). (b) First, we calculate the time-dependent current i ( t ) from Eq. 33-25: i ( t )= dq dt = d dt ³ Qe Rt/ 2 L sin( ω 0 t ) ´ = QR 2 L e Rt/ 2 L sin( ω 0 t )+ 0 e Rt/ 2 L cos( ω 0 t ) = Qe Rt/ 2 L µ R sin( ω 0 t ) 2 L + ω 0 cos( ω 0 t ) , which we evaluate at t =0 : i (0) = 0 .I fw ed en o t e i (0) = I as suggested in the problem, then Q = I/ω 0 . Returning this to Eq. 33-25 leads to q = Qe Rt/ 2 L cos( ω 0 t + φ )= µ I ω 0 e Rt/ 2 L cos ³ ω 0 t π 2 ´ =
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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