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Unformatted text preview: Physics 217: Problem Set 8 Solutions by Yao Liu Due Monday, April 13th, 2009 1 Purcell 8.7 To find the inductance of a toroid with N turn, see Problem 4 in Hw 6, B (2 r ) = 4 NI c = B = 2 NI cr = N Z b a Bhdr = 2 N 2 Ih c ln b a Definition of L is E =- L dI dt =- 1 c d dt so we read it off L = 2 N 2 h c 2 ln b a For the resonant cavity of this problem, N = 1. The capacitance of a parallel-plate capacitor is C = A 4 s = a 2 4 s = a 2 4 s where s is the gap. Therefore, the frequency of this LC circuit is = 1 LC = s 4 c 2 s 2 ha 2 ln( b/a ) = c a s 2 s h ln( b/a ) You should imagine E and B fields alternate. E field is in between the gap, and B is circumferential inside the toroid. That of course is not a complete picture; changing E field produces B field in the gap, and changing B field also produces E field in the toroid. The study of resonant cavities is complicated. Here you just get a flavor of it. 2 Purcell 8.8 I will explain as much as I can the relevant things about damped harmonic oscillator. This would be more appropriate in a course on analytical mechanics, where the typical context would be mass on a spring. The similarity between RLC circuit and damped mass-on-a-spring is manifest in the form of differential equations governing the two systems: L d 2 Q dt 2 + R dQ dt + Q C = 0 m d 2 x dt 2 + dx dt + kx = 0 As all ordinary differential equations go, we are solving it (the one on the left) subject to certain initial conditions. Since its second-order, we need two conditions. Typically we have Q (0) = Q , and Q (0) = 0, i.e., the capacitor is fully charged to Q , and there is no current going through the circuit. 1 2.1 Solution First step in solving a homogeneous linear o.d.e. with constant coefficients is to consider the characteristic equation (or secular equation) L 2 + R + 1 C = 0 which has two roots =- R p R 2- 4 L/C 2 L =- R 2 L r R 2 4 L 2- 1 LC =- q 2- 2 where = R/ 2 L is the damping factor, and = 1 / LC is the natural frequency of LC circuit. For purposes of this problem, wed like to think L and C as fixed, and we can change R at our will. In other words, is fixed, and we are tweaking damping . If the two roots are real and distinct ( > ), we say it is overdamped. The general solution is Q ( t ) = Ae 1 t + Be 2 t where A and B are determined from the two initial conditions. Note that both 1 and 2 are negative, which means Q ( t ) is exponential decay. Ill show you another way to write the solution, using hyperbolic functions defined by cosh x = e x + e- x 2 sinh x = e x- e- x 2 just like ordinary trigonometric functions can be defined as cos x = e ix + e- ix 2 sin x = e ix- e- ix 2 i Now the general solution can be written Q ( t ) = Ae- t cosh t + Be- t sinh t where = p 2- 2 (Naming it would make sense in the underdamped case). Imposing initial condition Q (0) = Q gives A = Q , and Q (0) = 0 gives A (- ) + B = 0, so...
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