solutionhw8 - Solutions to Homework Set 8 1 Pantograph Drag...

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Solutions to Homework Set 8 1) Pantograph Drag : The cable supports waves with dispersion equation ω 2 = Ω 2 + c 2 k 2 . Here c 2 = T/ρ . The phase velocity ω/k is always > c , while the group velocity ∂ω/∂k = c 2 k/ω is always < c . a) We seek a solution of the form y ( x - Ut ). Plugging this into the equation, we find that the function y ( x ) must obey - y 00 + γ 2 Ω 2 c 2 y = γ 2 F T δ ( x - Ut ) . Here γ 2 = (1 - U 2 /c 2 ) - 1 is positive, and so the solutions to the homogeneous equation are hyperbolic functions. The pantograph has no effect far from the train so we need to choose exponentially decaying solutions. Using the continuity and jump conditions at the delta function, we find. y ( x, t ) = γ 2 F T c γ ! exp - Ω γ c | x - Ut | . The overhead cable is pushed up symmetrically about the point of contact of the pantograph. F U U Cable profile for U < c As U approaches c from below γ will diverge to infinity, and our small-amplitude assumption will fail. b) We again look for a solution of the form y ( x - Ut ), but U > c and so y obeys - y 00 - ˜ γ 2 Ω 2 c 2 y = ˜ γ 2 F T δ ( x - Ut ) , where now it is ˜ γ 2 = ( U 2 /c 2 - 1) - 1 that is positive, and so the solutions to the homogeneous equations are oscillating functions. The locomotive is travelling faster than the group velocity of the waves. The cable is therefore unaware of the approaching train, and is undisturbed until the pantograph pick-up has passed. The solution will thus have y = 0 for x > Ut , and oscillate thereafter. The relevent solution is therefore y ( x, t ) = ( ˜ γ 2 F T c Ω˜ γ sin Ω˜ γ c ( Ut - x ) , x < Ut , 0 , x > Ut . This is the case illustrated by the drawing above the problem. As a check, observe that for this solution ω Ω˜ γU/c and k Ω˜ γ/c , obey the dispersion relation ω 2 = Ω 2 + c 2 k 2 , and so the wake is indeed a solution of the Klein-Gordon equation. 1
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c) The energy density in the wake is E = 1 2 ρ ˙ y 2 + 1 2 T ( y 0 ) 2 + 1 2 ρ Ω 2 y 2 = ρ ˜ γ 2 F T ! 2 U 2 sin 2 Ω˜ γ c | x - Ut | . The spatial average of sin 2 x is 1 / 2, so the average energy per unit length of wake is hEi = 1 2 ρ ˜ γ 4 F T 2 U 2 .
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