Explanation Given: The value of mass is 2 kg and the spring constant k is 32 N/m . Calculation: First convert the unit of spring constant ( k ) in Kg/sec 2 as follows. 1 N/m = 1 Kg/sec 2 32 N/m = 32 Kg/sec 2 The force acting on spring is expressed by a function as follows: f ( t ) = 68 e − 2 t cos 4 t The differential equation of forced motion is expressed as, m d 2 x d t 2 = − k x − β d x d t + f ( t ) (1) Substitute k = 32 and m = 2 in the equation (1) as follows: 2 ⋅ d 2 x d t 2 = − 32 x + 68 e − 2 t cos 4 t d 2 x d t 2 + 32 2 x = 68 2 e − 2 t cos 4 t d 2 x d t 2 + 16 x = 34 e − 2 t cos 4 t Therefore, the equation of motion is, d 2 x d t 2 + 16 x = 34 e − 2 t cos 4 t (2) As there is no force acting on the system the value of x ( 0 ) = 0 and at equilibrium, it is x ′ ( 0 ) = 0 . The homogeneous differential equation of equation (2) is, d 2 x d t 2 + 16 x = 0 (3) The auxiliary equation of equation (3) is, m 2 + 16 = 0 m 2 = − 16 m = ± 4 i Therefore, the root m 1 is − 4 i and m 2 is 4 i . The roots are complex then the solution is, x = e α t [ c 1 cos ( β t ) + c 2 sin ( β t ) ] Here, the value of α = 0 , β = 4 . And the complementary solution is, x c = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) (4) The particular solution is x p = e − 2 t ( A cos 4 t + B sin 4 t ) . Differentiate x p with respect to t . x ′ p = e − 2 t ( − 2 A cos 4 t − 2 B sin 4 t − 4 A sin 4 t + 4 B cos 4 t ) Now, differentiate x ′ p with respect to t. x ″ p ( t ) = − 2 e − 2 t ( − 2 A cos 4 t − 2 B sin 4 t − 4 A sin 4 t + 4 B cos 4 t ) + e − 2 t ( 8 A sin 4 t − 8 B cos 4 t − 16 A cos 4 t − 16 B sin 4 t ) = e − 2 t ( − 12 A cos 4 t − 12 B sin 4 t + 16 A sin 4 t − 16 B cos 4 t ) Substitute the values x p ( t ) , x ′ p ( t ) , x ″ p ( t ) in the given differential equation. e − 2 t ( 4 A cos 4 t + 4 B sin 4 t + 16 A sin 4 t − 16 B cos 4 t ) Compare the coefficients to both sides 4 A − 16 B = 34 16 A + 4 B = 0 Solve the above equations the value of A and B is, A = 1 2 and B = − 2 . Substitute the value of A and B in the particular solution x p = e − 2 t ( 1 2 cos 4 t − 2 sin 4 t ) Thus, the general solution of the differential equation x c + x p is, x = c 1 cos 4 t + c 2 sin 4 t + e − 2 t ( 1 2 cos 4 t − 2 sin 4 t ) (5) Differentiate the above equation with respect to t as, x ′ ( t ) = − 4 c 1 sin 4 t + 4 c 2 cos 4 t − 2 e − 2 t ( 1 2 cos 4 t − 2 sin 4 t ) + e − 2 t ( − 8 cos 4 t − 2 sin 4 t ) (6) The initial condition for the differential equation is x ( 0 ) = 0 and x ′ ( 0 ) = 0

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 5.1, Problem 38E

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To find: The equation of motion in the form of x(t)=Asin(ωt+ϕ)+Be−2tsin(4t+ϕ) and amplitude after very long time.

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Chapter 5 Solutions

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

Ch. 5.1 - 5.1.1 Spring/Mass systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A force...Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - 5.1.1Spring/Mass Systems: Free Undamped Motion A...

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The equation of motion in the form of x ( t ) = A sin ( ω t + ϕ ) + B e − 2 t sin ( 4 t + ϕ ) and amplitude after very long time.

Solution Summary: The author explains the equation of motion in the form of x(t)=Amathrmsin

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To find: The equation of motion in the form of x(t)=Asin(ωt+ϕ)+Be−2tsin(4t+ϕ) and amplitude after very long time.

Chapter 5 Solutions