Question

Consider the initial value problem my′'+ c y′+ k y=F(t),y(0)=0,y,(0)=0 modeling the motion of a spring-mass-dashpot system initially at rest and subjected to an applied force F(t),where the unit of force is the Newton (N). Assume that m=2 kilograms, c=8 kilograms per second, k= 80 Newtons per meter, and F(t)=20sin(6t)Newtons.a. Solve the initial value problem.b. Determine the long-term behavior of the system.c. Is limt→[infinity]y(t)=0?If no, enter a function that approximates y(t)for very large positive values of t.

Answer 1

Answer:

A)      $y_g = e^-^2^t*\frac{15}{37}cos(6t) + e^-^2^t*\frac{5}{74}sin(6t) + \frac{5}{74}sin(6t) - \frac{15}{37}cos(6t) \\\\y_g =\frac{15}{37}cos(6t)* [ e^-^2^t - 1 ] + \frac{5}{74}sin(6t)* [ e^-^2^t + 1 ]$

B)       $\frac{5}{74}sin(6t) - \frac{15}{37}cos(6t) = y_p$

Step-by-step explanation:

- The following initial value problem is given as follows:

                      $my'' + cy' + ky = F(t) \\\\y(0) = 0\\y'(0) = 0$

- The above equation is the Newtonian mathematical model of a spring-mass-dashpot system. The displacement ( y ) and velocity ( y' ) are zeroed at the initial value t = 0.

- The equivalent mass ( m ) , damping constant ( c ) and the equivalent spring stiffness ( k ) are given as follows:

                      $m = 2 kg\\\\c = 8 \frac{kg}{s} \\\\k = 80 \frac{N}{m}$

- The system is subjected to a sinusoidal force F ( t ) given. We will plug in the constants ( m , c, and k ) and applied force F ( t ) into the given second order ODE.

                      $2y'' + 8y' + 80y = 20sin(6t)$

- The solution to a second order ODE is comprised of a complementary function ( yc ) and particular function ( yp ).

- To determine the complementary function ( yc ) we will solve the homogeneous part of the given second order ODE. We will assume the independent solution to the homogeneous ODE takes the form:

                           $y = e^-^a^t$

Where,

                          a: The root of the following characteristic equation

- Substitute ( y ) into the given ODE as follows:

                          $( 2a^2 + 8a + 80 )*e^-^a^t = 0\\\\2a^2 + 8a + 80 = 0$

- Solve the above characteristic quadratic equation:

                           $a = 2 +/- 6i$

- The complementary solution for the complex solution to the characteristic equation is of the form:

                           $y_c = e^-^\alpha^t * [ Acos (\beta*t) + Bcos (\beta*t) ]$

Where,

                          a = α ± β

Therefore,

                           $y_c = e^-^2^t * [ Acos (6t) + Bcos (6t) ]$

- To determine the particular solution we will scrutinized on the non-homogeneous part of the given ODE. The forcing function F ( t ) the applied force governs the form of the particular solution. For sinusoidal wave-form the particular solution takes form as following:

                           $y_p = Csin (6t ) + Dcos(6t )$

Where,

                           C & D are constants to be evaluated.

- Determine the first and second derivatives of the particular solution (yp) as follows:

                            $y'_p = 6Ccos(6t) - 6Dsin(6t)\\\\y''_p = -36Ccos(6t) - 36Dcos(6t)$

- Plug in the particular solution ( yp ) and its derivatives ( first and second ) into the given ODE.

        $-72Csin(6t) - 72Dcos(6t) + 48Ccos(6t) - 48Dsin(6t) + 80Csin(6t) + 80Dcos(6t) = 20sin(6t) \\\\sin(6t)* ( 8C -48D ) + cos(6t)*(8D + 48C ) = 20sin(6t)\\\\D + 6C = 0\\\\C - 6D = 2.5\\\\C = \frac{5}{74} , D = -\frac{15}{37}$

- The particular solution can be written as follows:

                        $y_p = \frac{5}{74}sin(6t) - \frac{15}{37}cos(6t)$

- Now we use the principle of super-position and combine the complementary and particular solution and form a function of general solution as follows:

                        $y_g = y_c + y_p \\\\y_g = e^-^2^t* [ Acos(6t) + Bsin (6t) ] + \frac{5}{74}sin(6t) - \frac{15}{37}cos(6t)$

- To determine the complete solution of the given ODE we have to calculate the constants ( A and B ) using the given initial conditions as follows:

                        $y_g ( 0 ) = 1*[A(1) + 0 ] + 0 - \frac{15}{37}(1) = 0\\\\A = \frac{15}{37}\\\\y'_g = -2e^-^2^t*[Acos(6t) + Bsin(6t) ] +e^-^2^t*[-6Asin(6t) + 6Bcos(6t) ] + \\\\\frac{15}{37}cos(6t) +\frac{90}{37}sin(6t) \\\\y'_g(0) = -2*[A(1) + 0] + 1*[0 + 6B] + \frac{15}{37}(1) +0 = 0\\\\B = \frac{15}{6*37} = \frac{5}{74}$

- The complete solution to the initial value problem is:

                           

           $y_g = e^-^2^t*\frac{15}{37}cos(6t) + e^-^2^t*\frac{5}{74}sin(6t) + \frac{5}{74}sin(6t) - \frac{15}{37}cos(6t) \\\\y_g =\frac{15}{37}cos(6t)* [ e^-^2^t - 1 ] + \frac{5}{74}sin(6t)* [ e^-^2^t + 1 ]$          

- To determine the long term behavior of the system we will apply the following limit on our complete solution derived above:

                 $Lim (t->inf ) [ y_g ] = \frac{15}{37}cos(6t)* [ 0 - 1 ] + \frac{5}{74}sin(6t)* [ 0 + 1 ]\\\\Lim (t->inf ) [ y_g ] = \frac{5}{74}sin(6t) - \frac{15}{37}cos(6t) = y_p$

- We see that the complementary part of the solution decays as t gets large and the particular solution that models the applied force F ( t ) is still present in the system response when t gets large.