Explanation Rewriting the given force function, for t > τ consisting of two forcing functions F ( t ) m = [ 0 t < 0 a ( t τ ) 0 < t < τ a t > τ (I) Equation of motion for the interval 0 < t < τ is x ¨ + 2 β x ˙ + ω 0 2 x = a t τ (II) The particular solution is x p = C t + D . Differentiate the above relation with respect to t x ˙ = C Again differentiate the above relation with respect to t x ¨ = 0 Substitute the above solutions in equation (II) 2 β C + ω 0 2 C t + ω 0 2 D = a t τ (III) From equation (III) 2 β C + ω 0 2 D = 0 And, ω 0 2 C t − a t τ = 0 From above equation, ω 0 2 C t = a t τ C = a ω 0 2 τ Substitute the above equation in 2 β C + ω 0 2 D = 0 2 β ( a ω 0 2 τ ) + ω 0 2 D = 0 D = − 2 β a ω 0 4 τ ' Thus, the particular solution is x p = C t + D becomes, x p = a ω 0 2 τ t − 2 β a ω 0 4 τ ' The general solution for 0 < t < τ is x ( t ) = e − β t [ A cos ω 1 t + B sin ω 1 t ] + a ω 0 2 τ t − 2 β a ω 0 4 τ ' (IV) Differentiate equation (IV) with respect to t to solve for x ˙ x ˙ ( t ) = − β e − β t [ A cos ω 1 t + B sin ω 1 t ] + ω 1 e − β t [ − A sin ω 1 t + B cos ω 1 t ] + a ω 0 2 τ (V) For the initial conditions, x ( 0 ) = 0 then, x ˙ ( 0 ) = 0 Equation (IV) becomes, e 0 [ A cos 0 + B sin 0 ] + a ω 0 2 τ ( 0 ) − 2 β a ω 0 4 τ ' = 0 A − 2 β a ω 0 4 τ ' = 0 A = 2 β a ω 0 4 τ ' Equation (V) becomes, − β e 0 [ A cos 0 + B sin 0 ] + ω 1 e 0 [ − A sin 0 + B cos 0 ] + a ω 0 2 τ = 0 − β A + ω 1 B + a ω 0 2 τ = 0 − 2 β 2 a ω 0 4 τ ' + ω 1 B + a ω 0 2 τ = 0 ω 1 B = 2 β 2 a ω 0 4 τ ' − a ω 0 2 τ B = a ω 1 ω 0 2 τ ' ( 2 β 2 ω 0 2 − 1 ) The response function is x ( t ) = a ω 0 2 τ [ 2 β ω 0 2 e − β t cos ω 1 t +

Classical Dynamics of Particles and Systems

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Author: Stephen T. Thornton, Jerry B. Marion

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Chapter 3, Problem 3.31P

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The response function and show that for τ→0, the solution becomes the response for a step function.

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Chapter 3, Problem 3.30P

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Classical Dynamics of Particles and Systems

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