Explanation Given: The mass weight is 16 lbs and can stretch the spring 8 3 ft . It is also given that damping force is equal to the half of the instantaneous velocity. Calculation: By Hook’s law, m g = k s 16 = k × 8 3 k = 6 The mass m is, m = 16 32 = 1 2 slugs The equation of motion for a spring mass system having an external force is, m x ″ + β x ′ + k x = f ( t ) But, in the problem external force is given as f ( t ) = 10 cos 3 t . Substitute the values of m , k and β in the above equation: 1 2 x ″ + 1 2 x ′ + 6 x = 10 cos 3 t The auxiliary equation for the differential equation is, 1 2 m 2 + 1 2 m + 6 = 0 m 2 + m + 12 = 0 So, the values of m is, m = − 1 ± 1 2 − 4 × 1 × 12 2 = − 1 ± 1 − 48 2 = − 1 ± − 47 2 = − 1 ± i 47 2 Hence, the solution of the differential equation is, x ( t ) = e 1 2 t [ c 1 cos ( 47 2 t ) + c 2 sin ( 47 2 t ) ] . The particular solution and its derivative is, x p ( t ) = A cos 3 t + B sin 3 t x ′ p ( t ) = − 3 A sin 3 t + 3 B cos 3 t x ″ p ( t ) = − 9 A cos 3 t − 9 B sin 3 t Plugging these with differential equation, 1 2 ( − 9 A cos 3 t − 9 B sin 3 t ) + 1 2 ( − 3 A sin 3 t + 3 B cos 3 t ) + 6 ( A cos 3 t + B sin 3 t ) = 10 cos 3 t − 9 2 A cos 3 t − 9 2 B sin 3 t − 3 2 A sin 3 t − 3 2 B cos 3 t + 6 A cos 3 t + 6 B sin 3 t = 10 cos 3 t ( 3 2 A + 3 2 B ) cos 3 t + ( 3 2 B − 3 2 A ) sin 3 t = 10 cos 3 t Compare the coefficients of cos 3 t as follows

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3 Solutions

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Chapter 3 Solutions