Chem Differential Eq HW Solutions Fall 2011 25

Chem Differential Eq HW Solutions Fall 2011 25 - sin 3 t c...

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Section 2.7 Forced Oscillations 25 modifiedfurther t_ modifiedsteadystate t alph 1 Cos t bet 1 Sin t ; Plot Evaluate modifiedfurther t , steadystate t , modifiedsteadystate t , t, 0, 4 Pi 2 4 6 8 10 12 -0.4 -0.2 0.2 0.4 21. (a) The input function F ( t ) is already given by its Fourier series: F ( t )= 2 cos 2 t +sin 3 t . Since the frequency of the component sin 3 t of the input function is 3 and is equal to the natural frequency of the spring, resonance will occur (because there is no damping in the system). The general solution of y ±± +9 y = 2 cos 2 t +sin 3 t is y = y h + y p , where y h is the general solution of y ±± +9 y =0and y p is a particular solution of the nonhomogeneous equation. We have y h = c 1 sin 3 t + c 2 cos 3 t and, to ±nd y p , we apply Exercise 20 and get y p = ± a 2 A 2 cos 2 t + b 2 A 2 sin 2 t ² + R ( t ) , where a 2 =2 , b 2 =0 , A 2 =9 - 2 2 =5 , a n 0 =0 , b n 0 = 1, and R ( t )= - t 6 cos 3 t. Hence y p = 2 5 cos 2 t - t 6 cos 3 t and so the general solution is
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Unformatted text preview: sin 3 t + c 2 cos 3 t + 2 5 cos 2 t-t 6 cos 3 t. (b) To eliminate the resonance from the system we must remove the component of F that is causing resonance. Thus add to F ( t ) the function-sin 3 t . The modi±ed input function becomes F modi±ed ( t ) = 2 cos 2 t . 25. The general solution is y = c 1 sin 3 t + c 2 cos 3 t + 2 5 cos 2 t-t 6 cos 3 t . Applying the initial condition y (0) = 0 we get c 2 + 2 5 = 0 or c 2 =-2 5 . Thus y = c 1 sin 3 t-2 5 cos 3 t + 2 5 cos 2 t-t 6 cos 3 t. Applying the initial condition y ± (0) = 0, we obtain y ± = 3 c 1 cos 3 t + 6 5 sin 3 t-6 5 sin 2 t-1 6 cos 3 t + t 2 sin 3 t, y ± (0) = 3 c 1-1 6 , y ± (0) = 0 ⇒ c 1 = 1 18 ....
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