Explanation Given: The spring mass system is given by the differential equation as follows. m y ″ + b y ′ + k y = g ( t ) ... ( 1 ) Here, m > 0 is the mass of the spring system, b > 0 is the damping constant, k > 0 is the spring constant, g ( t ) is the force on the system at time t , and y ( t ) is the displacement from the equilibrium of the spring system at time t . The damping constant is b = 2 , the mass is m = 4 , the spring constant is k = 5 , and external force is g ( t ) = 2 sin 3 t + 10 cos 3 t . The boundary condition are given as y ( 0 ) = − 1 and y ′ ( 0 ) = 5 . Approach: The complementary solution for the differential equation if the roots of the auxiliary function are imaginary is given by, y c = c 1 e α t cos β t + c 2 e α t sin β t Here, the roots of the auxiliary equation are α + i β and α − i β and c 1 , c 2 are arbitrary constants. The general solution of the differential equation is given by, y = y c + y p Here, y p is the particular solution and y c is the complementary solution. Calculation: Substitute 1 for m , 5 for k , 2 for b , and ( 2 sin 3 t + 10 cos 3 t ) for g ( t ) in the equation ( 1 ) . ( 1 ) y ″ + ( 2 ) y ′ + ( 5 ) y = ( 2 sin 3 t + 10 cos 3 t ) y ″ + 2 y ′ + 5 y = 2 sin 3 t + 10 cos 3 t ... ( 2 ) The auxiliary equation of equation ( 2 ) is as follows. r 2 + 2 r + 5 = 0 Solve the above equation using the solution of quadratic equation. r 2 + 2 r + 5 = 0 r 2 + 2 r + 1 + 4 = 0 ( r + 1 ) 2 + 4 = 0 ( r + 1 ) 2 = − 4 Further solve, ( r + 1 ) = 2 i r = − 1 ± 2 i So the roots are, r 1 = − 1 + 2 i r 2 = − 1 − 2 i The complementary solution is as follows. y c = c 1 e − t cos ( 2 ) t + c 2 e − t sin ( 2 ) t Assume the particular solution of the differential equation as follows. y p = A sin 3 t + B cos 3 t ... ( 3 ) Differentiate the above equation with respect to t . y ′ = d d t ( A sin 3 t + B cos 3 t ) = 3 A cos 3 t − 3 B sin 3 t Again differentiate the above equation with respect to t . y ″ = d d t ( 3 A cos 3 t − 3 B sin 3 t ) = − 9 A sin 3 t − 9 B cos 3 t Substitute − 9 A sin 3 t − 9 B cos 3 t for y ″ , 3 A cos 3 t − 3 B sin 3 t for y ′ , and A sin 3 t + B cos 3 t for y in the equation ( 2 ) . ( − 9 A sin 3 t − 9 B cos 3 t + 2 ( 3 A cos 3 t − 3 B sin 3 t ) + 5 ( A sin 3 t + B cos 3 t ) ) = 2 sin 3 t + 10 cos 3 t ( − 9 A sin 3 t − 9 B cos 3 t + 6 A cos 3 t − 6 B sin 3 t + 5 A sin 3 t + 5 B cos 3 t ) = 2 sin 3 t + 10 cos 3 t ( − 9 A − 6 B + 5 A ) cos 3 t + ( − 9 B + 6 A + 5 B ) sin 3 t = 2 sin 3 t + 10 cos 3 t ( − 4 A − 6 B ) cos 3 t + ( 6 A − 4 B ) sin 3 t = 2 sin 3 t + 10 cos 3 t ... ( 4 ) Compare the coefficients of sin 3 t on both sides in above equation. − 4 A − 6 B = 2 4 A = − 2 − 6 B A = − 2 − 6 B 4 A = − 1 − 3 B 2

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

See similar textbooks

Videos

Question

Chapter 4.5, Problem 44E

To determine

To find:

The equation of motion for the spring system.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.5, Problem 43E

Chapter 4.5, Problem 45E

Students have asked these similar questions

5. The volume V of a given mass of monoatomic gas changes with temperat re T according to the relation V = KT2/3. The work done when temperature changes by 90 K will be xR. The value of x is (a) 60 (b)20 (c)30 S (d)90

Consider a matrix 3 -2 1 A = 0 5 4 -6 2 -1 Define matrix B as transpose of the inverse of matrix A. Find the determinant of matrix A + B.

For each of the time series, construct a line chart of the data and identify the characteristics of the time series (that is, random, stationary, trend, seasonal, or cyclical). Year Month Rate (%)2009 Mar 8.72009 Apr 9.02009 May 9.42009 Jun 9.52009 Jul 9.52009 Aug 9.62009 Sep 9.82009 Oct 10.02009 Nov 9.92009 Dec 9.92010 Jan 9.82010 Feb 9.82010 Mar 9.92010 Apr 9.92010 May 9.62010 Jun 9.42010 Jul 9.52010 Aug 9.52010 Sep 9.52010 Oct 9.52010 Nov 9.82010 Dec 9.32011 Jan 9.12011 Feb 9.02011 Mar 8.92011 Apr 9.02011 May 9.02011 Jun 9.12011 Jul 9.02011 Aug 9.02011 Sep 9.02011 Oct 8.92011 Nov 8.62011 Dec 8.52012 Jan 8.32012 Feb 8.32012 Mar 8.22012 Apr 8.12012 May 8.22012 Jun 8.22012 Jul 8.22012 Aug 8.12012 Sep 7.82012 Oct…

Chapter 4 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

Ch. 4.1 - Verify that for b=0 and Fext(t)=0, equation (3)...Ch. 4.1 - If Fext(t)=0, equation (3) becomes my+by+ky=0. For...Ch. 4.1 - Show that if Fext(t)=0, m=1, k=9, and b=6, then...Ch. 4.1 - Prob. 4E Ch. 4.1 - Verify that the exponentially damped sinusoid...Ch. 4.1 - An external force F(t)=2cos2t is applied to a...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Undamped oscillators that are driven at resonance...

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.

The equation of motion for the spring system.

Solution Summary: The author explains how the equation of motion for the spring system is given by the differential equation.

Videos

To find: The equation of motion for the spring system.

Want to see the full answer?

Chapter 4 Solutions

To find:

The equation of motion for the spring system.