# ssm_ch31 - Chapter 31 7(a The mass m corresponds to the...

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Chapter31 7 (a) The mass m corresponds to the inductance, so m = 1 : 25 kg. (b) The spring constant k corresponds to the reciprocal of the capacitance. Since the total energy is given by U = Q 2 = 2 C , where Q is the maximum charge on the capacitor and C is the capacitance, C = Q 2 2 U = ¡ 175 £ 10 ¡ 6 C ¢ 2 2(5 : 70 £ 10 ¡ 6 J) = 2 : 69 £ 10 ¡ 3 F and k = 1 2 : 69 £ 10 ¡ 3 m = N = 372 N = m : (c) The maximum displacement x m corresponds to the maximum charge, so x m = 1 : 75 £ 10 ¡ 4 m : (d) The maximum speed v m corresponds to the maximum current. The maximum current is I = Q! = Q p LC = 175 £ 10 ¡ 6 C p (1 : 25 H)(2 : 69 £ 10 ¡ 3 F) = 3 : 02 £ 10 ¡ 3 A : Thus v m = 3 : 02 £ 10 ¡ 3 m = s. 15 (a) Since the frequency of oscillation f is related to the inductance L and capacitance C by f = 1 = 2 ¼ p LC , the smaller value of C gives the larger value of f . Hence, f max = 1 = 2 ¼ p LC min , f min = 1 = 2 ¼ p LC max , and f max f min = p C max p C min = p 365 pF p 10 pF = 6 : 0 : (b) You want to choose the additional capacitance C so the ratio of the frequencies is r = 1 : 60 MHz 0 : 54 MHz = 2 : 96 : Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that of the tuning capacitor. If C is in picofarads, then p C + 365 pF p C + 10 pF = 2 : 96 : 194 Chapter 31

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The solution for C is C = (365 pF) ¡ (2 : 96) 2 (10 pF) (2 : 96) 2 ¡ 1 = 36 pF : (c) Solve f = 1 = 2 ¼ p LC for L . For the minimum frequency, C = 365 pF + 36 pF = 401 pF and f = 0 : 54 MHz. Thus L = 1 (2 ¼ ) 2 Cf 2 = 1 (2 ¼ ) 2 (401 £ 10 ¡ 12 F)(0 : 54 £ 10 6 Hz) 2 = 2 : 2 £ 10 ¡ 4 H : 27 Let t be a time at which the capacitor is fully charged in some cycle and let q max 1 be the charge on the capacitor then. The energy in the capacitor at that time is U ( t ) = q 2 max 1 2 C = Q 2 2 C e ¡ Rt=L ; where q max 1 = Q e ¡ Rt= 2 L was used. Here Q is the charge at t = 0. One cycle later, the maximum charge is q max 2 = Q e ¡ R ( t + T ) = 2 L and the energy is U ( t + T ) = q 2 max 2 2 C = Q 2 2 C e ¡ R ( t + T ) =L ; where T is the period of oscillation. The fractional loss in energy is ¢ U U = U ( t ) ¡ U ( t + T ) U ( t ) = e ¡ Rt=L ¡ e ¡ R ( t + T ) =L e ¡ Rt=L = 1 ¡ e ¡ RT=L : Assume that RT=L is small compared to 1 (the resistance is small) and use the Maclaurin series to expand the exponential. The first two terms are: e ¡ RT=L ¼ 1 ¡ RT L : Replace T with 2 ¼=! , where ! is the angular frequency of oscillation. Thus ¢ U U ¼ 1 ¡ μ 1 ¡ RT L = RT L = 2 ¼R !L : 33 (a) The generator emf is a maximum when sin( ! d t ¡ ¼= 4) = 1 or ! d t ¡ ¼= 4 = ( ¼= 2) § 2 , where n is an integer, including zero. The first time this occurs after t = 0 is when ! d t ¡ ¼= 4 = ¼= 2 or t = 3 ¼ 4 ! d = 3 ¼ 4(350 s ¡ 1 ) = 6 : 73 £ 10 ¡ 3 s : (b) The current is a maximum when sin( ! d t ¡ 3 ¼= 4) = 1, or ! d t ¡ 3 ¼= 4 = ¼= 2 § 2 . The first time this occurs after t = 0 is when t = 5 ¼ 4 !
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