c11mudd - 18.03 Muddy Card responses March 3 2006 1 I...

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18.03 Muddy Card responses, March 3, 2006 1. I confused a number of people by dividing through m ¨ x + bx ˙ + kx = F ext ( t ) and ¨ miraculously getting x + bx ˙ + kx = q ( t ). What I meant to say was that by dividing through you make the coefficient of x equal to 1 and replace the others by b/m and ¨ k/m ; but then I’ll change the meaning of the symbol b so that it means the damping constant divided by the mass, and of k so that it means the spring constant divided by the mass. 2. Here’s the proof that if x 1 and x 2 are solutions of the homogeneous, linear second ¨ order ODE x + b ( t ) ˙ x + k ( t ) x = 0 then so is x = c 1 x 1 + c 2 x 2 . Plug this x into the equation: k ( t ) x = k ( t ) c 1 x 1 + k ( t ) c 2 x 2 b ( t ) ˙ x = b ( t ) c 1 x ˙ 1 + k ( t ) c 2 x ˙ 2 x ¨ ¨ ¨ = c 1 x 1 + c 2 x 2 ¨ x 1 + b ( t ) ˙ x 1 + k ( t ) x 1 ) + c 2 x + b ( t ) ˙ x + k ( t ) x = c 1 x 2 + b ( t ) ˙ x 2 + b ( t ) x 2 ) and both terms on the right are zero, by assumption. As the notation indicates, this
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.

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