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

Bundle: A First Course in Differential Equations with Modeling Applications, Loose-leaf Version, 11th + WebAssign Printed Access Card for Zill's ... Problems, 9th Edition, Single-Term

11th Edition

ISBN: 9781337761000

Author: Dennis G. Zill

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 5.1, Problem 38E

To determine

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.

Expert Solution & Answer

Chapter 5 Solutions

Bundle: A First Course in Differential Equations with Modeling Applications, Loose-leaf Version, 11th + WebAssign Printed Access Card for Zill's ... Problems, 9th Edition, Single-Term

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

Recommended textbooks for you

Trigonometry (MindTap Course List)
Trigonometry
ISBN:9781337278461
Author:Ron Larson
Publisher:Cengage Learning

Trigonometry (MindTap Course List)

Trigonometry

ISBN:9781337278461

Author:Ron Larson

Publisher:Cengage Learning

SEE MORE TEXTBOOKS

But what is the Fourier Transform? A visual introduction.; Author: 3Blue1Brown;https://www.youtube.com/watch?v=spUNpyF58BY;License: Standard YouTube License, CC-BY

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

Videos

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