Answer to Question #147626 in Differential Equations for Walaa

Question

1- Suppose a body weighing 64 Ibs attached to a spring stretches It by 8 feet . Assume that damping force, numerically equal to 8 times the instantaneous velocity acts on the mass .In addition an external force f(t)=6t2 4 is being applied to the system .At t=0 the body is released from rest at appoint 3 feet above the equilibrium position.
a- Setup a mathematical model to find displacement of the body from equilibrium position at time t.
b- Solve the model analytically and find a formula for the displacement at any time t.
C- Draw simulation block diagram.

Answer 1

Note : All calculations will be done in SI.

It means that

$$\left\{\begin{array}{l} m=64\,\,lbs\approx 29.03\,kg\\[0.3cm] \Delta x=8\,\,feet\approx 2.44\,\,m\\[0.3cm] l_0=3\,\,feet\approx 0.92\,\,m\\[0.3cm] g\approx 9.81\,\,m/s^2 \end{array}\right.$$

1 STEP : Find the stiffness of the spring

(more information : https://en.wikipedia.org/wiki/Hooke%27s_law)

In our case,

$$mg=k\cdot\Delta x\to k=\frac{mg}{\Delta x}\approx\frac{29.03\cdot9.81}{2.44}\approx116.72\,\frac{N}{m}$$

2 STEP : Let's describe all the forces that act on our body

$$\text{The force of gravity} - F_g=mg\\[0.3cm] \text{Hooke's force}- F_s=-kx\\[0.3cm] \text{damping force}-F_d=-8\cdot v\equiv-8\cdot\frac{dx}{dt}\\[0.3cm] \text{an external force}- f(t)=6t^2$$

3 STEP : Let's write the equation of motion (Newton's second law) for a given body, together with the initial conditions

$$m\vec{a}=\vec{F_g}+\vec{F_s}+\vec{F_d}+\vec{f}(t)\\[0.3cm] x(0)=l_0\approx0.92\,m\\[0.3cm] x'(0)=0$$

Suppose that all movement occurs in the vertical direction, the axis is directed along the action of the force of gravity. Here is the corresponding figure.

Note: since the forces of Hooke, dumping and one more act in the same direction, for the convenience of reading the diagram, I will draw from NOT from one point.Since the direction of the external force is not indicated, then I myself will choose its direction.

$$Oy : m\cdot\overbrace{a}^{x''}=mg-8\overbrace{v}^{x'}-116.72\cdot x+6t^2\\[0.3cm] 29.03\cdot x''+8\cdot x'+116.72\cdot x=29.03\cdot 9.81+6t^2$$

This is an inhomogeneous second-order equation with constant coefficients. The solution has the form

$$x(t)=x_{hom}(t)+x_{part}^{(1)}+x_{part}^{(2)},\,\,\text{where}\\[0.3cm] x_{hom}(t) : 29.03\cdot x''+8\cdot x'+116.72\cdot x=0\\[0.3cm] x(t)=e^{\lambda t}\to x'(t)=\lambda\cdot e^{\lambda t}\to x''(t)=\lambda^2\cdot e^{\lambda t}\\[0.3cm] \left.29.03\cdot \lambda^2\cdot e^{\lambda t}+8\cdot \lambda\cdot e^{\lambda t}+116.72\cdot e^{\lambda t}=0\right|\div e^{\lambda t}\\[0.3cm] 29.03\cdot \lambda^2+8\cdot \lambda+116.72=0\\[0.3cm] \sqrt{D}=\sqrt{8^2-4\cdot29.03\cdot116.72}\approx 116.14\cdot i\\[0.3cm] \left[\begin{array}{l} \lambda_1=\displaystyle\frac{-8-116.14\cdot i}{2\cdot29.09}\approx-0.14-2i\\[0.3cm] \lambda_2=\displaystyle\frac{-8+116.14\cdot i}{2\cdot29.09}\approx-0.14+2i \end{array}\right.\\[0.3cm] x_{hom}(t)=A_1\cdot e^{(-0.14-2i)t}+A_2\cdot e^{(-0.14+2i)t}\to\\[0.3cm] \boxed{x_{hom}(t)=C_1e^{-0.14t}\cdot\cos{2t}+C_2e^{-0.14t}\cdot\sin{2t}}$$

It remains to find partial solutions :

$$x_{part}^{(1)} : 29.03\cdot x''+8\cdot x'+116.72\cdot x=29.03\cdot 9.81\\[0.3cm] x(t)=Const\to x'(t)=0\to x''(t)=0\to\\[0.3cm] 29.03\cdot 0+8\cdot 0+116.72\cdot Const=29.03\cdot 9.81\to\\[0.3cm] Const=\frac{29.03\cdot 9.81}{116.72}\approx2.44\to\boxed{x_{part}^{(1)}=2.44}\\[0.3cm] x_{part}^{(2)} : 29.03\cdot x''+8\cdot x'+116.72\cdot x=6t^2\\[0.3cm] x(t)=At^2+Bt+C\to x'(t)=2At+b\to x''(t)=2A\to\\[0.3cm] 29.03\cdot 2A+8\cdot(2At+B)+116.72\cdot(At^2+Bt+C)=6t^2\to\\[0.3cm] 116.72At^2+(116.72B+16A)t+(116.72C+8B+48.06A)=6t^2\\[0.3cm] \left\{\begin{array}{l} 116.72A=6\\[0.3cm] 116.72B+16A=0\\[0.3cm] 116.72C+8B+48.06A=0 \end{array}\right.\to \left\{\begin{array}{l} A\approx 0.05\\[0.3cm] B\approx-0.007\\[0.3cm] C\approx-0.02 \end{array}\right.\\[0.3cm] \boxed{x_{part}^{(2)}=0.05t^2-0.007t-0.02}$$

Conclusion, general solution is

$$x(t)=x_{hom}(t)+x_{part}^{(1)}+x_{part}^{(2)}\\[0.3cm] x(t)=e^{-0.14t}\cdot\left(C_1\cos{2t}+C_2\sin{2t}\right)+0.05t^2-0.007t+2.42$$

Now let's find the constants, for this we use the initial conditions:

$$x(0)\approx0.92=1\cdot(C_1\cdot1+C_2\cdot0)+2.42\to\\[0.3cm] C_1=0.92-2.42=-1.5\to\boxed{C_1=-1.5}\\[0.3cm] x'(t)=-0.14e^{-0.14t}\cdot\left(C_1\cos{2t}+C_2\sin{2t}\right)+\\[0.3cm] +e^{-0.14t}\cdot\left(-2C_1\sin{2t}+2C_2\cos{2t}\right)+0.1t-0.007\\[0.3cm] x'(0)=0=-0.14\cdot C_1+2C_2-0.007\to\\[0.3cm] 2C_2=0.14\cdot1.5+0.007=0.217\to \boxed{C_2=0.1085}$$

Final result :

$$\boxed{x(t)=e^{-0.14t}\left(-1.5\cos(2t)+0.1085\sin(2t)\right)+0.05t^2-0.007t+2.42}$$

ANSWER :

(a) model

$$29.03\cdot x''+8\cdot x'+116.72\cdot x=29.03\cdot 9.81+6t^2\\[0.3cm] x(0)=l_0\approx0.92\,m\\[0.3cm] x'(0)=0$$

(b) solution

$$x(t)=e^{-0.14t}\left(-1.5\cos(2t)+0.1085\sin(2t)\right)+0.05t^2-0.007t+2.42$$

(c) graph

For small times $0\le t\le 10$

For large times $0\le t\le 50$