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Unformatted text preview: 80. (a) From Eq. 3325, RQ −Rt/2L d dq cos(ω t + φ) − ω Qe−Rt/2L sin(ω t + φ) = Qe−Rt/2L cos(ω t + φ) = − e dt dt 2L and d2 q dt2 = R 2L e−Rt/2L RQ cos(ω t + φ) − ω Q sin(ω t + φ) 2L RQω sin(ω t + φ) − ω 2 Q cos(ω t + φ) . 2L + e−Rt/2L Substituting these expressions, and Eq. 3325 itself, into Eq. 3324, we obtain Qe−Rt/2L − ω 2 L − R 2L
2 + 1 cos(ω t + φ) = 0 . c Since this equation is valid at any time t, we must have −ω 2 L − R 2L
2 + 1 = 0 =⇒ ω = C 1 − LC R 2L 2 = ω2 − R 2L 2 . (b) The fractional shift in frequency is ∆ω ∆f = f ω = ω−ω =1− ω 1− 1− (1/LC ) − (R/2L)2 1/LC =1− 1− R2 C 4L = (100 Ω)2 (7.30 × 10−6 F) = 0.00210 = 0.210% . 4(4.40 H) ...
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 Fall '08
 SPRUNGER
 Physics

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