Mathematics: Consider the initial value problem my +cy +ky=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)=20 sin(6t) Newtons. 1. Solve the initial value problem. y(t)=? 2. Determine the long term behavior of the system. Is lim as t goes to infinity of y(t)=0? If it is, enter zero. If not, enter a

The answer is the second one from the top so the 1/3 with the 8 on the bottom.Step-by-step explanation: No...

Consider the initial value problem my +cy +ky=F(t),

Ответ:

bella4269

13.06.2021

Hence, the $y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t))\end{array}$ and approximately value of y(t) is -0.844 .

Given :

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

Where m=2 kilograms

c=8 kilograms per second

k=80 Newtons per meter

$F(t)=20\sin (6t)$ Newtons

Explanation :

(1)

Solve the initial value problem. y(t)

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

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

$\Rightarrow y''+4y'+40y=10\sin (6t)$

Auxilary equations : F(t)=0

$\Rightarrow r^2+4r+40=0$

$\Rightarrow r=\frac{-4\pm\sqrt{4^2-4\times 1\times 40}}{2\times 1}$

$\Rightarrow r=\frac{-4\pm\sqrt{16-160}}{2}$

$\Rightarrow r=\frac{-4\pm\sqrt{-144}}{2}$

$\Rightarrow r=\frac{-4\pm12i}{2}$

$\Rightarrow r=-2\pm6i$

The complementary solution is $y_c=e^{-2t}\left(c_1\cos 6t+c_2\sin 6t\right)$

The particular Integral, $y_p=\frac{1}{f(D)}F(t)$

$y_{y} &=\frac{1}{D^{2}+4 D+40} 25 \sin (6 t) \\\\ y_{y} &=\frac{25}{-6^{2}+4 D+40} \sin (6 t) \quad\left(D^{2} \text { is replaced with }-6^{2}=-36\right) \\\\y_{y} &=\frac{25}{4 D+4} \sin (6 t) \\\\y_{y} &=\frac{25}{4(D+1)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4(D+1)(D-1)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4\left(D^{2}-1\right)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4(-36-1)} \sin (6 t) \\\\y_{y} &=-\frac{25}{148}(D-1) \sin (6 t) \\y_{y} &=-\frac{25}{148}\left(\frac{d}{d t} \sin (6 t)-\sin (6$

Hence the general solution is : $y=y_c+y_p=e^{-2t}(c_1\cos 6t+c_2\sin 6t)-\frac{25}{148}(6\cos 6t-\sin 6t)$

Now we use given initial condition.

$y(t) &=e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\\y(0) &=e^{-\alpha 0)}\left(c_{1} \cos (0)+c_{2} \sin (0)\right)-\frac{25}{148}(6 \cos (0)-\sin (0)) \\\\0 &=\left(c_{1}\right)-\frac{25}{148}(6) \\\\c_{1} &=\frac{75}{74} \\\\y(t) &=e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\$

$y^{\prime}(t)=-2 e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)+e^{-2 t}\left(-6 c_{1} \sin 6 t+6 c_{2} \cos 6 t\right)-\frac{25}{148}(-36 \sin (6 t)-6 \cos (6 t)) \\\\y^{\prime}(0)=-2 e^{0}\left(c_{1} \cos 0+c_{2} \sin 0\right)+e^{0}\left(-6 c_{1} \sin 0+6 c_{2} \cos 0\right)-\frac{25}{148}(-36 \sin 0-6 \cos 0) \\\\0=-2\left(c_{1}\right)+\left(6 c_{2}\right)-\frac{25}{148}(-6) \\\\0=-2 c_{1}+6 c_{2}+\frac{75}{74} \\\\0=-2\left(\frac{75}{74}\right)+6 c_{2}+\frac{75}{74} \\\\$ $\begin{array}{l}0=-\frac{150}{74}+6 c_{2}+\frac{75}{74} \\\\\frac{150}{74}-\frac{75}{74}=6 c_{2}\end{array}$

$\begin{array}{l}c_{2}=\frac{25}{148}\\\\\text { Substitute } c_{1} \text { and } c_{2} \text { in } y(t) \text { . Then }\\\\y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t))\end{array}$

(2)

$y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\left(\frac{75}{74} e^{-2 t} \cos 6 t+\frac{75}{148} e^{-2 t} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\frac{75}{74}\left(e^{-2 t}-1\right) \cos 6 t+\frac{25}{148}\left(3 e^{-2 t}+1\right) \sin 6 t \\\\|y(t)| \leq \frac{75}{74} e^{-2 t}-1|\cos 6 t|+\frac{25}{148}\left|3 e^{-2 t}+1\right||\sin 6 t| \\\\$

$|y(t)| \leq \frac{75}{74}\left|e^{-2 t}-1\right|+\frac{25}{148}\left|3 e^{-2 t}+1\right| \\\\\lim _{t \rightarrow \infty} y(t) \leq \lim _{t \rightarrow \infty}\left\{\frac{75}{74}\left|e^{-2 t}-1\right|+\frac{25}{148}\left|3 e^{-2 t}+1\right|\right\} \\\\\lim _{t \rightarrow \infty} y(t)=\left\{\frac{75}{74}\left|\lim _{t \rightarrow \infty}\left(e^{-2 t}-1\right)\right|+\frac{25}{148}\left|\lim _{t \rightarrow \infty}\left(3 e^{-2 t}+1\right)\right|\right\} \\$

$\lim _{t \rightarrow \infty} y(t)=\left\{\frac{75}{74}(-1)+\frac{25}{148}(1)\right\}=-\frac{75}{74}+\frac{25}{148}=-\frac{-150+25}{148}=-\frac{125}{148} \approx-0.844$

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question