190_pdfsam_math 54 differential equation solutions odd

190_pdfsam_math 54 differential equation solutions odd - c...

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Chapter 4 (a) We want to determine the equation of motion for a spring system with m =10kg , b = 60 kg/sec, k = 250 kg/sec 2 , y (0) = 0 . 3m ,and y 0 (0) = 0 . 1 m/sec. That is, we seek the solution to the initial value problem 10 y 0 ( t )+60 y 0 ( t ) + 250 y ( t )=0; y (0) = 0 . 3 ,y 0 (0) = 0 . 1 . The auxiliary equation for the above diFerential equation is 10 r 2 +60 r + 250 = 0 r 2 +6 r +25=0 , which has the roots r = 6 ± 36 100 2 = 6 ± 8 i 2 = 3 ± 4 i. Hence α = 3and β = 4, and the displacement y ( t ) has the form y ( t )= c 1 e 3 t cos 4 t + c 2 e 3 t sin 4 t. We ±nd c 1 and c 2 by using the initial conditions. We ±rst diFerentiate y ( t )toget y 0 ( t )=( 3 c 1 +4 c 2 ) e 3 t cos 4 t +( 4 c 1 3 c 2 ) e 3 t sin 4 t. Substituting y and y 0 into the initial conditions, we obtain the system y (0) = 0 . 3= c 1 , y 0 (0) = 0 . 1= 3 c 1 +4 c 2 .
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Unformatted text preview: c 1 = 0 . 3 and c 2 = 0 . 2. Therefore the equation of motion is given by y ( t ) = 0 . 3 e − 3 t cos 4 t + 0 . 2 e − 3 t sin 4 t (m) . (b) ²rom Problem 32 we know that the frequency of oscillation is given by β/ (2 π ). In part (a) we found that β = 4. Therefore the frequency of oscillation is 4 / (2 π ) = 2 /π . (c) We see a decrease in the frequency of oscillation. We also have the introduction of the factor e − 3 t , which causes the solution to decay to zero. This is a result of energy loss due to the damping. 186...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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