886_PartUniversity Physics Solution

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

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30-10 Chapter 30 (c) SET UP: At critical damping, 4/ R LC = . EXECUTE: () 3 9 422 .0 10 H 2420 15.0 10 F R × == Ω × EVALUATE: 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. IDENTIFY: Follow the procedure specified in the problem. SET UP: Make the substitutions 1 ,, , xq mL bR k C →→→→ . EXECUTE: (a) Eq. (13.41): 2 2 0 dx bd x k x dt m dt m ++ = . This becomes 2 2 0 dq Rdq q dt L dt LC + += , which is Eq.(30.27). (b) Eq. (13.43): 2 2 4 kb ω mm =− . This becomes 2 2 1 4 R ω L CL , which is Eq.(30.29). (c) Eq. (13.42): (/ 2 ) cos( ) bm t xA e ω t φ =+ . This becomes 2) cos( ) RL t qA e ω t = + , which is Eq.(30.28). EVALUATE: Equations for the L - R - C circuit and for a damped harmonic oscillator have the same form. 30.40. IDENTIFY: For part (a), evaluate the derivatives as specified in the problem. For part (b) set qQ = in Eq.(30.28) and set /0 dq dt = in the expression for / dq dt . SET UP: In terms of ω , Eq.(30.28) is cos ( ) t qt Ae t = + . EXECUTE: (a) cos( ) t e ω t . cos( ) sin( ). 2 t t dq R ω t ω Ae ω t dt L −− ′′ + + 2 2 2 2 cos( ) 2 sin( ) cos( ) 22 t t t dq R R ω t ω ω t ω ω t dt L L φφ ⎛⎞ + + + ⎜⎟ ⎝⎠ 2 2 1 0 q R R q dt L dt LC L L LC = , so 2 2 2 1 4 R ω L . (b) At 0, , 0 dq tq Q i dt === = , so cos Q and cos sin 0 2 dq R A ω A dt L = = . This gives cos Q A = and tan 2 21 / / 4 RR L ω L LC R L . EVALUATE: If 0 R = , then AQ = and 0 = . 30.41. IDENTIFY: Evaluate Eq.(30.29). SET UP: The angular frequency of the circuit is . EXECUTE: (a) When 0 5 11 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 (1 4 ) 1( 0 . 9 5 ) 14 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 × EVALUATE: When R increases, the angular frequency decreases and approaches zero as 2/ R . 30.42. IDENTIFY: L has units of H and C has units of F. SET UP: 1 H 1 s . / CqV = says 1 F 1 C/V. = VI R = says 1 V/A 1 = Ω . EXECUTE: The units of / L C are 2 HsV . FC V A Ω⋅ = Ω Therefore, the unit of / L C is . Ω EVALUATE: 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. IDENTIFY: The emf 2 E in solenoid 2 produced by changing current 1 i in solenoid 1 is given by 1 2 . i M t Δ = Δ E The mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem 01 2 , AN N M l μ = where A is the cross-sectional area of the inner solenoid and l is the length of the outer solenoid.

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Inductance 30-11 SET UP: 7 0 41 0 T m / A μπ . Let the outer solenoid be solenoid 1.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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