Program Explanation Given information: The value of mass and spring constants are written below. m 1 = 1 , m 2 = 1 2 k 1 = 75 , k 2 = 25 , k 3 = 0 The walls are not exist for the given spring mass system. The stiffness matrix for the given mass and spring system is written as, K = [ − ( k 1 + k 2 ) k 2 k 2 − ( k 2 + k 3 ) ] The given external forces F 1 ( t ) and F 2 ( t ) acting on the masses m 1 and m 2 respectively as, F 1 ( t ) = 0 F 2 ( t ) = 100 cos 10 t The initial conditions for the system is written as, x ( 0 ) = 0 x ′ ( 0 ) = 0 Explanation: The matrix differential equation for the given system is written as, M x ″ = K x + F Here the mass matrix M is written as, M = [ m 1 0 0 m 2 ] M = [ 2 0 0 1 2 ] Similarly the stiffness matrix K is written as, K = [ − ( k 1 + k 2 ) k 2 k 2 − ( k 2 + k 3 ) ] K = [ − ( 75 + 25 ) 25 25 − ( 25 + 0 ) ] K = [ − 100 25 25 − 25 ] Similarly the external force vector F is written as, F = [ F 1 ( t ) F 2 ( t ) ] F = [ 0 100 cos 10 t ] F = [ 0 100 ] cos 10 t Therefore, matrix differential equation for the given system is written as, [ 2 0 0 1 2 ] x ″ = [ − 100 25 25 − 25 ] x + [ 0 100 ] cos 10 t Reduce the above differential equation in simplest form as, x ″ = [ 2 0 0 1 2 ] − 1 [ − 100 25 25 − 25 ] x + [ 2 0 0 1 2 ] − 1 [ 0 100 ] cos 10 t x ″ = [ 0.5 0 0 2 ] [ − 100 25 25 − 25 ] x + [ 0.5 0 0 2 ] [ 0 100 ] cos 10 t x ″ = [ − 50 12.5 50 − 50 ] x + [ 0 200 ] cos 10 t Consider the periodic particular solution as x p ( t ) = c cos 10 t and so derivatives are written as, x ′ p ( t ) = − 10 c sin 10 t x ″ p ( t ) = − 100 c cos 10 t Substitute, the above periodic particular solution and its derivative in the matrix differential equation and solve for the vector c as, − 100 c cos 10 t = [ − 50 12.5 50 − 50 ] c cos 10 t + [ 0 200 ] cos 10 t − 100 c = [ − 50 12.5 50 − 50 ] c + [ 0 200 ] − [ 0 200 ] = [ − 50 12.5 50 − 50 ] c + 100 c [ 0 − 200 ] = [ 50 12.5 50 50 ] c Further, solve the above equation as, [ 0 − 200 ] = [ 50 12.5 50 50 ] [ c 1 c 2 ] [ 0 − 200 ] = [ 50 c 1 + 12.5 c 2 50 c 1 + 50 c 2 ] { 50 c 1 + 12.5 c 2 = 0 50 c 1 + 50 c 2 = − 200 Solve the system of equations by substitution method and obtain the value of c 1 and c 2 as, c 1 = 4 3 c 2 = − 16 3 Therefore, the period solution x p ( t ) = c cos 10 t of the equation is written as, x p ( t ) = c cos 10 t x p ( t ) = [ c 1 c 2 ] cos 10 t x p ( t ) = [ 4 3 − 16 3 ] cos 10 t Solve the homogeneous part of the non-homogeneous equation as follows: The matrix A for the homogeneous part is written below

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

See similar textbooks

Concept explainers

Rod Cutting

A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

Question

Chapter 5.4, Problem 15P

Program Plan Intro

Program Description: Purpose of the problem is to find the solution of the system Mx″=Kx+F for the initial conditions x(0)=x′(0)=0 .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5.4, Problem 14P

Chapter 5.4, Problem 16P

Students have asked these similar questions

For an object of mass m=3 kg to slide without friction up the rise of height h=1 m shown, it must have a minimum initial kinetic energy (in J) of: h O a. 40 O b. 20 O c. 30 O d. 10

Q.4 In an experimental setup, mineral oil is filled in between the narrow gap of two horizontal smooth plates. The setup has arrangements to maintain the plates at desired uniform temperatures. At these temperatures, ONLY the radiative heat flux is negligible. The thermal conductivity of the oil does not vary perceptibly in this temperature range. Consider four experiments at steady state under different experimental conditions, as shown in the figure Q1. The figure shows plate temperatures and the heat fluxes in the vertical direction. What is the steady state heat flux (in W m) with the top plate at 90°C and the bottom plate at 45°C? [4] 30°C 70°C 40°C 90°C flux = 39 Wm-2 flux =30 Wm2 flux = 52 Wm 2 flux ? Wm-2 60°C 35°C 80°C 45°C Experiment 1 Experiment 2 Experiment 3 Experiment 4

I need the answer as soon as possible

Chapter 5 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

Ch. 5.1 - Prob. 11P Ch. 5.1 - Prob. 12P Ch. 5.1 - Prob. 13P Ch. 5.1 - Prob. 14P Ch. 5.1 - Prob. 15P Ch. 5.1 - Prob. 16P Ch. 5.1 - Prob. 17P Ch. 5.1 - Prob. 18P Ch. 5.1 - Prob. 19P Ch. 5.1 - Prob. 20P Ch. 5.1 - Prob. 21P Ch. 5.1 - Prob. 22P Ch. 5.1 - Prob. 23P Ch. 5.1 - Prob. 24P Ch. 5.1 - Prob. 25P Ch. 5.1 - Prob. 26P Ch. 5.1 - Prob. 27P Ch. 5.1 - Prob. 28P Ch. 5.1 - Prob. 29P Ch. 5.1 - Prob. 30P Ch. 5.1 - Prob. 31P Ch. 5.1 - Prob. 32P Ch. 5.1 - Prob. 33P Ch. 5.1 - Prob. 34P Ch. 5.1 - Prob. 35P Ch. 5.1 - Prob. 36P Ch. 5.1 - Prob. 37P Ch. 5.1 - Prob. 38P Ch. 5.1 - Prob. 39P Ch. 5.1 - Prob. 40P Ch. 5.1 - Prob. 41P Ch. 5.1 - Prob. 42P Ch. 5.1 - Prob. 43P Ch. 5.1 - Prob. 44P Ch. 5.1 - Prob. 45P Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.2 - Prob. 10P Ch. 5.2 - Prob. 11P Ch. 5.2 - Prob. 12P Ch. 5.2 - Prob. 13P Ch. 5.2 - Prob. 14P Ch. 5.2 - Prob. 15P Ch. 5.2 - Prob. 16P Ch. 5.2 - Prob. 17P Ch. 5.2 - Prob. 18P Ch. 5.2 - Prob. 19P Ch. 5.2 - Prob. 20P Ch. 5.2 - Prob. 21P Ch. 5.2 - Prob. 22P Ch. 5.2 - Prob. 23P Ch. 5.2 - Prob. 24P Ch. 5.2 - Prob. 25P Ch. 5.2 - Prob. 26P Ch. 5.2 - Prob. 27P Ch. 5.2 - Prob. 28P Ch. 5.2 - Prob. 29P Ch. 5.2 - Prob. 30P Ch. 5.2 - Prob. 31P Ch. 5.2 - Prob. 32P Ch. 5.2 - Prob. 33P Ch. 5.2 - Prob. 34P Ch. 5.2 - Prob. 35P Ch. 5.2 - Prob. 36P Ch. 5.2 - Prob. 37P Ch. 5.2 - Prob. 38P Ch. 5.2 - Prob. 39P Ch. 5.2 - Prob. 40P Ch. 5.2 - Prob. 41P Ch. 5.2 - Prob. 42P Ch. 5.2 - Prob. 43P Ch. 5.2 - Prob. 44P Ch. 5.2 - Prob. 45P Ch. 5.2 - Prob. 46P Ch. 5.2 - Prob. 47P Ch. 5.2 - Prob. 48P Ch. 5.2 - Prob. 49P Ch. 5.2 - Prob. 50P Ch. 5.3 - Prob. 1P Ch. 5.3 - Prob. 2P Ch. 5.3 - Prob. 3P Ch. 5.3 - Prob. 4P Ch. 5.3 - Prob. 5P Ch. 5.3 - Prob. 6P Ch. 5.3 - Prob. 7P Ch. 5.3 - Prob. 8P Ch. 5.3 - Prob. 9P Ch. 5.3 - Prob. 10P Ch. 5.3 - Prob. 11P Ch. 5.3 - Prob. 12P Ch. 5.3 - Prob. 13P Ch. 5.3 - Prob. 14P Ch. 5.3 - Prob. 15P Ch. 5.3 - Prob. 16P Ch. 5.3 - Prob. 17P Ch. 5.3 - Prob. 18P Ch. 5.3 - Prob. 19P Ch. 5.3 - Prob. 20P Ch. 5.3 - Prob. 21P Ch. 5.3 - Prob. 22P Ch. 5.3 - Prob. 23P Ch. 5.3 - Prob. 24P Ch. 5.3 - Prob. 25P Ch. 5.3 - Prob. 26P Ch. 5.3 - Prob. 27P Ch. 5.3 - Prob. 28P Ch. 5.3 - Prob. 29P Ch. 5.3 - Prob. 30P Ch. 5.3 - Prob. 31P Ch. 5.3 - Prob. 32P Ch. 5.3 - Prob. 33P Ch. 5.3 - Verify Eq. (53) by substituting the expressions...Ch. 5.3 - Prob. 35P Ch. 5.3 - Prob. 36P Ch. 5.3 - Prob. 37P Ch. 5.3 - Prob. 38P Ch. 5.3 - Prob. 39P Ch. 5.3 - Prob. 40P Ch. 5.4 - Prob. 1P Ch. 5.4 - Prob. 2P Ch. 5.4 - Prob. 3P Ch. 5.4 - Prob. 4P Ch. 5.4 - Prob. 5P Ch. 5.4 - Prob. 6P Ch. 5.4 - Prob. 7P Ch. 5.4 - Prob. 8P Ch. 5.4 - Prob. 9P Ch. 5.4 - Prob. 10P Ch. 5.4 - Prob. 11P Ch. 5.4 - Prob. 12P Ch. 5.4 - Prob. 13P Ch. 5.4 - Prob. 14P Ch. 5.4 - Prob. 15P Ch. 5.4 - Prob. 16P Ch. 5.4 - Prob. 17P Ch. 5.4 - Prob. 18P Ch. 5.4 - Prob. 19P Ch. 5.4 - Prob. 20P Ch. 5.4 - Prob. 21P Ch. 5.4 - Prob. 22P Ch. 5.4 - Prob. 23P Ch. 5.4 - Prob. 24P Ch. 5.4 - Prob. 25P Ch. 5.4 - Prob. 26P Ch. 5.4 - Prob. 27P Ch. 5.4 - Prob. 28P Ch. 5.4 - Prob. 29P Ch. 5.5 - Prob. 1P Ch. 5.5 - Prob. 2P Ch. 5.5 - Prob. 3P Ch. 5.5 - Prob. 4P Ch. 5.5 - Prob. 5P Ch. 5.5 - Prob. 6P Ch. 5.5 - Prob. 7P Ch. 5.5 - Prob. 8P Ch. 5.5 - Prob. 9P Ch. 5.5 - Prob. 10P Ch. 5.5 - Prob. 11P Ch. 5.5 - Prob. 12P Ch. 5.5 - Prob. 13P Ch. 5.5 - Prob. 14P Ch. 5.5 - Prob. 15P Ch. 5.5 - Prob. 16P Ch. 5.5 - Prob. 17P Ch. 5.5 - Prob. 18P Ch. 5.5 - Prob. 19P Ch. 5.5 - Prob. 20P Ch. 5.5 - Prob. 21P Ch. 5.5 - Prob. 22P Ch. 5.5 - Prob. 23P Ch. 5.5 - Prob. 24P Ch. 5.5 - Prob. 25P Ch. 5.5 - Prob. 26P Ch. 5.5 - Prob. 27P Ch. 5.5 - Prob. 28P Ch. 5.5 - Prob. 29P Ch. 5.5 - Prob. 30P Ch. 5.5 - Prob. 31P Ch. 5.5 - Prob. 32P Ch. 5.5 - Prob. 33P Ch. 5.5 - Prob. 34P Ch. 5.5 - Prob. 35P Ch. 5.5 - Prob. 36P Ch. 5.6 - Prob. 1P Ch. 5.6 - Prob. 2P Ch. 5.6 - Prob. 3P Ch. 5.6 - Prob. 4P Ch. 5.6 - Prob. 5P Ch. 5.6 - Prob. 6P Ch. 5.6 - Prob. 7P Ch. 5.6 - Prob. 8P Ch. 5.6 - Prob. 9P Ch. 5.6 - Prob. 10P Ch. 5.6 - Prob. 11P Ch. 5.6 - Prob. 12P Ch. 5.6 - Prob. 13P Ch. 5.6 - Prob. 14P Ch. 5.6 - Prob. 15P Ch. 5.6 - Prob. 16P Ch. 5.6 - Prob. 17P Ch. 5.6 - Prob. 18P Ch. 5.6 - Prob. 19P Ch. 5.6 - Prob. 20P Ch. 5.6 - Prob. 21P Ch. 5.6 - Prob. 22P Ch. 5.6 - Prob. 23P Ch. 5.6 - Prob. 24P Ch. 5.6 - Prob. 25P Ch. 5.6 - Prob. 26P Ch. 5.6 - Prob. 27P Ch. 5.6 - Prob. 28P Ch. 5.6 - Prob. 29P Ch. 5.6 - Prob. 30P Ch. 5.6 - Prob. 31P Ch. 5.6 - Prob. 32P Ch. 5.6 - Prob. 33P Ch. 5.6 - Prob. 34P Ch. 5.6 - Prob. 35P Ch. 5.6 - Prob. 36P Ch. 5.6 - Prob. 37P Ch. 5.6 - Prob. 38P Ch. 5.6 - Prob. 39P Ch. 5.6 - Prob. 40P Ch. 5.7 - Prob. 1P Ch. 5.7 - Prob. 2P Ch. 5.7 - Prob. 3P Ch. 5.7 - Prob. 4P Ch. 5.7 - Prob. 5P Ch. 5.7 - Prob. 6P Ch. 5.7 - Prob. 7P Ch. 5.7 - Prob. 8P Ch. 5.7 - Prob. 9P Ch. 5.7 - Prob. 10P Ch. 5.7 - Prob. 11P Ch. 5.7 - Prob. 12P Ch. 5.7 - Prob. 13P Ch. 5.7 - Prob. 14P Ch. 5.7 - Prob. 15P Ch. 5.7 - Prob. 16P Ch. 5.7 - Prob. 17P Ch. 5.7 - Prob. 18P Ch. 5.7 - Prob. 19P Ch. 5.7 - Prob. 20P Ch. 5.7 - Prob. 21P Ch. 5.7 - Prob. 22P Ch. 5.7 - Prob. 23P Ch. 5.7 - Prob. 24P Ch. 5.7 - Prob. 25P Ch. 5.7 - Prob. 26P Ch. 5.7 - Prob. 27P Ch. 5.7 - Prob. 28P Ch. 5.7 - Prob. 29P Ch. 5.7 - Prob. 30P Ch. 5.7 - Prob. 31P Ch. 5.7 - Prob. 32P Ch. 5.7 - Prob. 33P Ch. 5.7 - Prob. 34P

Knowledge Booster

Learn more about

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

Program Description: Purpose of the problem is to find the solution of the system M x ″ = K x + F for the initial conditions x ( 0 ) = x ′ ( 0 ) = 0 .

Solution Summary: The author explains how the matrix differential equation for the given system is written as, Mx′′=Kx+F.

Concept explainers

Program Description: Purpose of the problem is to find the solution of the system Mx″=Kx+F for the initial conditions x(0)=x′(0)=0 .

Want to see the full answer?

Chapter 5 Solutions