Program Explanation Given information: The differential equation is, m x ″ + c x ′ + k x = E 0 cos ω t + F 0 sin ω t Explanation: Let x s p ( t ) = x s p 1 ( t ) + x s p 2 ( t ) . Therefore, calculate the separate steady periodic solution corresponding to E 0 cos ω t and F 0 sin ω t . Consider the equation of motion as, x ( t ) = A cos ω t + B sin ω t Differentiate the above equation with respect to t as, x ′ ( t ) = − A ω sin ω t + B ω cos ω t Again, differentiate the above equation with respect to t as, x ″ ( t ) = − A ω 2 cos ω t − B ω 2 sin ω t Substitute the values of x , x ′ a n d x ″ in the equation m x ″ + c x ′ + k x = E 0 cos ω t as, m ( − A ω 2 cos ω t − B ω 2 sin ω t ) + c ( − A ω sin ω t + B ω cos ω t ) + k ( A cos ω t + B sin ω t ) = E 0 cos ω t ( c ω A + ( k − m ω 2 ) B ) sin ω t + ( ( k − m ω 2 ) A + c ω B ) cos ω t = E 0 cos ω t Compare the coefficients in the above equation as, ( k − m ω 2 ) A + c ω B = E 0 − c ω A + ( k − m ω 2 ) B = 0 ⇒ A = ( k − m ω 2 ) E 0 ( k − m ω 2 ) 2 + ( c w ) 2 ⇒ B = c ω E 0 ( k − m ω 2 ) 2 + ( c w ) 2 Substitute the value of A and B in the equation x ( t ) = A cos ω t + B sin ω t as, x ( t ) = ( k − m ω 2 ) E 0 ( k − m ω 2 ) 2 + ( c w ) 2 cos ω t + c ω E 0 ( k − m ω 2 ) 2 + ( c w ) 2 sin ω t The above equation can be written as, − A cos ω t + B sin ω t = ( − A C cos ω t + B C sin ω t ) C The value of C is calculated as, C = ( ( k − m ω 2 ) E 0 ( k − m ω 2 ) 2 + ( c w ) 2 ) 2 + ( c ω E 0 ( k − m ω 2 ) 2 + ( c w ) 2 ) 2 C = E 0 ( k − m ω 2 ) 2 + c 2 ω 2 Therefore, x s p 1 ( t ) = E 0 ( k − m ω 2 ) 2 + c 2 ω 2 cos ( ω t − α ) . Now, the differential equation is m x ″ + c x ′ + k x = F 0 sin ω t . Consider the equation of motion as, x ( t ) = A cos ω t + B sin ω t Differentiate the above equation with respect to t as, x ′ ( t ) = − A ω sin ω t + B ω cos ω t Again, differentiate the above equation with respect to t as, x ″ ( t ) = − A ω 2 cos ω t − B ω 2 sin ω t Substitute the values of x , x ′ a n d x ″ in the equation m x ″ + c x ′ + k x = F 0 sin ω t as, [ m ( − A ω 2 cos ω t − B ω 2 sin ω t ) + c ( − A ω sin ω t + B ω cos ω t ) + k ( A cos ω t + B sin ω t ) ] = F 0 sin ω t ( c ω A + ( k − m ω 2 ) B ) sin ω t + ( ( k − m ω 2 ) A + c ω B ) cos ω t = F 0 sin ω t Compare the coefficients in the above equation as, ( k − m ω 2 ) A + c ω B = 0 ⇒ A = c ω B ( k − m ω 2 ) − c ω A +

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 3.6, Problem 29P

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Program Description: Purpose of problem is to show that the amplitude C of the resulting steady periodic oscillation for the car is C=ak2+ ( cω )2 ( k−m ω 2 )2+ ( cω )2

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Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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Program Description: Purpose of problem is to show that the amplitude C of the resulting steady periodic oscillation for the car is C = a k 2 + ( c ω ) 2 ( k − m ω 2 ) 2 + ( c ω ) 2

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Program Description: Purpose of problem is to show that the amplitude C of the resulting steady periodic oscillation for the car is C=ak2+ ( cω )2 ( k−m ω 2 )2+ ( cω )2

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