323_pdfsam_math 54 differential equation solutions odd

# 323_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 5.6 To ﬁnd the steady-state current, we must, therefore, ﬁnd A and B . To accomplish this, we observe that Ip (t) = −20A sin 20t + 20B cos 20t, Ip (t) = −400A cos 20t − 400B sin 20t. Plugging these expressions into equation (5.36) yields Ip (t) + 30Ip (t) + 550I (t) = −400A cos 20t − 400B sin 20t − 600A sin 20t + 600B cos 20t +550A cos 20t + 550B sin 20t = −50 sin 20t ⇒ (150A + 600B ) cos 20t + (150B − 600A) sin 20t = −50 sin 20t. By equating coeﬃcients we obtain the system of equations 15A + 60B = 0, −60A + 15B = −5. By solving these equations simultaneously for A and B , we obtain A = 4/51 and B = −1/51. Thus, we have the steady-state current given by 1 4 cos 20t − sin 20t. Ip (t) = 51 51 As was observed on page 290 of the text, there is a correlation between the RLC series circuits and mechanical vibration. Therefore, we can discuss the resonance frequency of the RLC series circuit. To do so we associate the variable L with m, R with b, and 1/C with k . Thus, we see that the resonance frequency for an RLC series circuit is given by γr /(2π ), where γr = provided that R2 < 2L/C . For this problem R2 = 14, 400 < 2L/C = 17, 600 . Therefore, we can ﬁnd the resonance frequency of this circuit. To do so we ﬁrst ﬁnd R2 1 2200 14400 − 2= − = 10. CL 2L 4 32 Hence the resonance frequency of this circuit is 10/(2π ) = 5/π . γr = 319 R2 1 − 2, CL 2L ...
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