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prelim1999 - Table of Contents Complete Partially Complete...

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Table of Contents Complete Partially Complete Not Complete Preliminary Examination 1999 2 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1
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Preliminary Examination September , 1999 1. Consider a yo-yo climbing its string. Assume that the string is vertical. Notation: r = radius of yo-yo axle m = mass of yo-yo I = moment of inertia of yo-yo g = acceleration of gravity ω = angular velocity of yo-yo about its axis Show that the equations of motion for the yo-yo reduce to an equation for ˙ ω in terms of r , m , I , and g (and nothing else). Show that your equation implies that the total energy of the yo-yo is conserved as it climbs. Why must that be true? Solution (a) Incomplete: Need to answer energy question.] y th r PSfrag replacements θ y r The kinetic energy of our yo-yo is given by T = 1 2 m ˙ y 2 + 1 2 I ˙ θ 2 Because positive y is in the downward direction, the potential energy is V = mgy Thus, the Lagrangian is L = 1 2 m ˙ y 2 + 1 2 I ˙ θ 2 + mgy Also, we note that our system is under the following constraint: f ( y, θ ) = y = 0 ˙ y = r ˙ θ ¨ y = r ¨ θ Thus, the Euler-Lagrange equations are ∂L ∂y d d t ∂L ˙ y + λ ∂f ∂y = 0 mg m ¨ y + λ = 0 λ = m ( g ¨ y ) ∂L ∂θ d d t ∂L ˙ θ + λ ∂f ∂θ = 0 I ¨ θ + λ ( r ) = 0 I ¨ θ + mr ( g ¨ y ) = 0 ¨ θ ( I + mr 2 ) = mgr ¨ θ = ˙ ω = mgr I + mr 2 2
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Preliminary Examination September , 1999 2. The scalar and vector potentials of an oscillating dipole with dipole moment p 0 and frequency ω are respectively: V ( r, θ, t ) = ( p 0 ω 4 πϵ 0 c ) ( cos θ r ) sin [ ω ( t r c )] and A ( r, θ, t ) = ( µ 0 p 0 ω 4 πr ) sin [ ω ( t r c )] ˆ k (a) Show that these potentials are in the Lorentz gauge. (b) Find the electric and magnetic fields and the Poynting vector. (c) Find the total power radiated. (d) Speculate why the sky is blue. 3
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Preliminary Examination September , 1999 3. Consider the partial differential equation t ∂f ∂t + 2 f ∂x 2 = 0 Find a solution for t 1 , 0 x 2 , subject to the following boundary conditions f ( x, 1) = 1 0 x 1 2 2(1 x ) 1 2 x 3 2 1 3 2 x 2 ∂f ∂x (0 , t ) = ∂f ∂x (2 , t ) = 0 (a) Use separation of variables to find two corresponding ordinary differential equations. (b) What are the solutions to the ODE’s from part a? You need to give only the solutions relevant for the boundary conditions, although if you prefer you may give all solutions. (c) Do the Fourier expansion appropriate to the boundary conditions. (You do not need to evaluate trig functions appearing in the Fourier coefficients). (d) Combine the above to give the complete solution to the equation. Solution (a) Let f ( x, t ) = F ( x ) G ( t ) . The partial differential equation becomes tF d G d t + G d 2 F d x 2 = 0 t G d G d t = 1 F d 2 F d x 2 = k 2 Thus, our ordinary differential equations are d 2 F d x 2 = k 2 F t d G d t = k 2 G (b) First, let us find the solution to the x equation.
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