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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 = 10 kg, b = 60 kg/sec, k = 250 kg/sec 2 , y (0) = 0 . 3 m, and y (0) = 0 . 1 m/sec. That is, we seek the solution to the initial value problem 10 y ( t ) + 60 y ( t ) + 250 y ( t ) = 0; y (0) = 0 . 3 , y (0) = 0 . 1 . The auxiliary equation for the above differential 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 = 3 and 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 find c 1 and c 2 by using the initial conditions. We first differentiate y ( t ) to get y ( 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 into the initial conditions, we obtain the system y (0) = 0 . 3 = c 1 , y (0) = 0 . 1 = 3 c 1
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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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