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886_PartUniversity Physics Solution

886_PartUniversity Physics Solution - 30-10 Chapter 30(c...

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30-10 Chapter 30 (c) S ET U P : At critical damping, 4 / R L C = . E XECUTE : ( ) 3 9 4 22.0 10 H 2420 15.0 10 F R × = = Ω × E VALUATE : The frequency with damping is almost the same as the resonance frequency of this circuit ( 1/ LC ), which is plausible because the 75- resistance is considerably less than the 2420 required for critical damping. 30.39. I DENTIFY : Follow the procedure specified in the problem. S ET U P : Make the substitutions 1 , , , x q m L b R k C . E XECUTE : (a) Eq. (13.41): 2 2 0 d x b dx kx dt m dt m + + = . This becomes 2 2 0 d q R dq q dt L dt LC + + = , which is Eq.(30.27). (b) Eq. (13.43): 2 2 4 k b ω m m ′ = . This becomes 2 2 1 4 R ω LC L ′ = , which is Eq.(30.29). (c) Eq. (13.42): ( / 2 ) cos( ) b m t x Ae ω t φ = + . This becomes ( / 2 ) cos( ) R L t q Ae ω t φ = + , which is Eq.(30.28). E VALUATE : Equations for the L - R - C circuit and for a damped harmonic oscillator have the same form. 30.40. I DENTIFY : For part (a), evaluate the derivatives as specified in the problem. For part (b) set q Q = in Eq.(30.28) and set / 0 dq dt = in the expression for / dq dt . S ET U P : In terms of ω , Eq.(30.28) is ( / 2 ) ( ) cos( ) R L t q t Ae t ω φ = + . E XECUTE : (a) ( / 2 ) cos( ) R L t q Ae ω t φ = + . ( / 2 ) ( / 2 ) cos( ) sin( ). 2 R L t R L t dq R A e ω t ω Ae ω t dt L φ φ = − + + 2 2 ( / 2 ) ( / 2 ) 2 ( / 2 ) 2 cos( ) 2 sin( ) cos( ) 2 2 R L t R L t R L t d q R R A e ω t ω A e ω t ω Ae ω t dt L L φ φ φ = + + + + 2 2 2 2 2 2 1 0 2 2 d q R dq q R R q dt L dt LC L L LC ω + + = + = , so 2 2 2 1 4 R ω LC L ′ = . (b) At 0, , 0 dq t q Q i dt = = = = , so cos q A Q φ = = and cos sin 0 2 dq R A ω A dt L φ φ = − = . This gives cos Q A φ = and 2 2 tan 2 2 1/ /4 R R L ω L LC R L φ = − = − . E VALUATE : If 0 R = , then A Q = and 0 φ = . 30.41. I DENTIFY : Evaluate Eq.(30.29). S ET U P : The angular frequency of the circuit is ω . E XECUTE : (a) When 0 5 1 1 0, 298 rad s. (0.450 H) (2.50 10 F) R ω LC = = = = × (b) We want 0 0.95 ω ω = , so 2 2 2 2 (1 4 ) 1 (0.95) 1 4 LC R L R C LC L = = . This gives 2 5 4 4(0.450 H)(0.0975) (1 (0.95) ) 83.8 . (2.50 10 F) L R C = = = Ω × E VALUATE : When R increases, the angular frequency decreases and approaches zero as 2 / R L C . 30.42. I DENTIFY : L has units of H and C has units of F. S ET U P : 1 H 1 s = Ω⋅ . / C q V = says 1 F 1 C/V. = V IR = says 1 V/A 1 = Ω . E XECUTE : The units of / L C are 2 H s V . F C V A Ω⋅ Ω⋅ = = = Ω Therefore, the unit of / L C is . Ω E VALUATE : For Eq.(30.28) to be valid, 1 LC and 2 2 4 R L must have the same units, so R and / L C must have the same units, and we have shown that this is indeed the case. 30.43. I DENTIFY : The emf 2 E in solenoid 2 produced by changing current 1 i in solenoid 1 is given by 1 2 .
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