237_pdfsam_math 54 differential equation solutions odd

# 237_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.8 It follows that period = 2 π ω = 2 π 4 = π 2 , natural frequency = ω 2 π = 2 π . A general solution to (4.10), given in (4) on page 211 in the text, becomes y ( t )= C 1 cos ωt + C 2 sin ωt = C 1 cos 4 t + C 2 sin 4 t. We Fnd C 1 and C 2 from the initial conditions. y (0) = ( C 1 cos 4 t + C 2 sin 4 t ) ± ± t =0 = C 1 = 1 / 2 , y 0 (0) = ( 4 C 1 sin 4 t +4 C 2 cos 4 t ) ± ± t =0 =4 C 2 =2 C 1 = 1 / 2 , C 2 =1 / 2 . Thus, the solution to the initial value problem is y ( t )= 1 2 cos 4 t + 1 2 sin 4 t = 2 2 sin ² 4 t π 4 ³ , where we have used formulas (6) rewriting the solution in form (5), page 211 in the text. The amplitude of the motion therefore is 2 / 2. Setting y = 0 in the above solution, we Fnd values of t when the mass passes through the point of equilibrium.
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