A 2 kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 20 cm upon coming to rest at equilibrium. At time t = 0, the mass is displaced 5 cm below the equilibrium position and released. At this same instant, an external force F(t) = 0.3 cos t N is applied to the systems. If the damping constant for the system is 5 N-sec/m, determine the equation of the motion for the mass. What is the resonance frequency for the system?

The general solution to (1) in the case $0<{\mathrm{b}}^2<4\mathrm{mk}$:

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\left(\frac{\mathrm{b}}{2\mathrm{m}}\right)\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}\sin \left(\gamma \mathrm{t}+\mathrm{\theta }\right)$

The angular frequency:

The amplitude of the steady–state solution to equation (1) depends on the angular frequency$y$ of the forcing function and it is given by $\mathrm{A}\left(\gamma \right)={\mathrm{F}}_0\mathrm{M}\left(\gamma \right)$, where

(13)$\mathrm{M}\left(\gamma \right):=\frac{1}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}$ … (1)

The undamped system:

The system is governed by $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \gamma \mathrm{t}$. And the homogenous solution of it is${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right),\omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$is . And the corresponding homogeneous equation is

(21). And the correct form is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos}\omega \mathrm{t}+{\mathrm{A}}_2\mathrm{tsin}\omega \mathrm{t}$.

So, the general solution to the system is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos}\omega \mathrm{t}+{\mathrm{A}}_2\mathrm{tsin}\omega \mathrm{t}$.

Given that, $\mathrm{m}=2$. And $\mathrm{F}=0.3\mathrm{cost}$. So,${\mathrm{F}}_0=0.3$ and $\gamma =1$. Since $\mathrm{b}=5$, and the length of the stretch is 20 cm. Use Hooke’s law to find the value of k.

$$\begin{gathered} \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}} \\ \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}} \\ =\frac{2\times 9.8}{0.2}=98 \end{gathered}$$

Then, the differential equation is $2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+5\frac{\mathrm{dy}}{\mathrm{dt}}+98\mathrm{y}=0.3\cos \left(\mathrm{t}\right)$.

Since,

$$\begin{gathered} {\mathrm{b}}^2=25 \\ 4\mathrm{mk}=784 \end{gathered}$$

One knows that, ${\mathrm{b}}^2<4\mathrm{mk}$so we are dealing with the underdamped motion.

The homogeneous solution is given by

$$\begin{gathered} {\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\varphi \right) \\ ={\mathrm{Ae}}^{-\frac{5}{4}\mathrm{t}}\sin \left(\frac{\sqrt{784-25}}{4}\mathrm{t}+\varphi \right) \\ ={\mathrm{Ae}}^{-\frac{5}{4}\mathrm{t}}\sin \left(\frac{\sqrt{759}}{4}\mathrm{t}+\varphi \right) \end{gathered}$$

Where A and$\varphi$ are constants.

Then, the particular solution is given by

$$\begin{gathered} {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}\sin \left(\gamma \mathrm{t}+\mathrm{\theta }\right) \\ =\frac{0.3}{\sqrt{{\left(98-2\right)}^2+25}}\sin \left(\mathrm{t}+\mathrm{\theta }\right) \\ =\frac{3}{10\sqrt{9241}}\sin \left(\mathrm{t}+\mathrm{\theta }\right) \end{gathered}$$

The general solution is$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\frac{5}{4}\mathrm{t}}\sin \left(\frac{\sqrt{759}}{4}\mathrm{t}+\varphi \right)+\frac{3}{10\sqrt{9241}}\sin \left(\mathrm{t}+\mathrm{\theta }\right)$ … (2)

At t = 0, the mass is at 0.05 m, and at rest, that is$\mathrm{y}\left(0\right)=0.05,\mathrm{y}\text{'}\left(0\right)=0$ , .

Now find the derivative of equation (2).

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{A}\left(-\frac{5}{4}{\mathrm{e}}^{-\frac{5}{4}\mathrm{t}}\sin \left(\frac{\sqrt{759}}{4}\mathrm{t}+\varphi \right)+\frac{\sqrt{759}}{4}{\mathrm{e}}^{-\frac{5}{4}\mathrm{t}}\cos \left(\frac{\sqrt{759}}{4}\mathrm{t}+\varphi \right)\right)+\frac{3}{10\sqrt{9241}}\cos \left(\mathrm{t}+\mathrm{\theta }\right)$

Now find the constants using the initial conditions.

$$\begin{gathered} \mathrm{y}\left(0\right)={\mathrm{Ae}}^{-0}\sin \left(0+\varphi \right)+\frac{3}{10\sqrt{9241}}\sin \left(0+\mathrm{\theta }\right) \\ 0.05=\mathrm{Asin}\left(\varphi \right)+\frac{3}{10\sqrt{9241}}\sin \left(\mathrm{\theta }\right) \end{gathered}$$

So,$\mathrm{Asin}\left(\varphi \right)=0.05-\frac{3}{10\sqrt{9241}}\sin \left(\mathrm{\theta }\right)$ … (3)

$$\begin{gathered} \left(0\right)=\mathrm{A}\left(-\frac{5}{4}{\mathrm{e}}^{-0}\sin \left(0+\varphi \right)+\frac{\sqrt{759}}{4}{\mathrm{e}}^{-0}\cos \left(0+\varphi \right)\right)+\frac{3}{10\sqrt{9241}}\cos \left(0+\mathrm{\theta }\right) \\ 0=-\frac{5}{4}\mathrm{Asin}\left(\varphi \right)+\frac{\sqrt{759}}{4}\mathrm{Acos}\left(\varphi \right)+\frac{3}{10\sqrt{9241}}\cos \left(\mathrm{\theta }\right) \end{gathered}$$

Now solve the equation (3)

$$\begin{gathered} \mathrm{Asin}\left(\varphi \right)=0.05-\frac{3}{10\sqrt{9241}}\sin \left(\mathrm{\theta }\right) \\ \mathrm{A}=\frac{0.05-\frac{3}{10\sqrt{9241}}\sin \left({\tan }^{-1}\left(\frac{96}{5}\right)\right)}{\sin \left(\varphi \right)} \\ =\frac{1733}{36964\sin \left(\varphi \right)} \end{gathered}$$

Then, find the value of A.

Using the fact of $\sin \left({\cot }^{-1}\left(\mathrm{x}\right)\right)=\frac{1}{\sqrt{{\mathrm{x}}^2+1}}$.

$$\begin{gathered} \mathrm{A}=\frac{1733}{36964\sin \left({\cot }^{-1}\left(\frac{8641\sqrt{759}}{1315347}\right)\right)} \\ =\sqrt{\frac{15922}{7013919}} \end{gathered}$$

So, the solution is:

$\mathrm{y}\left(\mathrm{t}\right)=\sqrt{\frac{15922}{7013919}}{\mathrm{e}}^{-\frac{5}{4}\mathrm{t}}\sin \left(\frac{\sqrt{759}}{4}\mathrm{t}+{\cot }^{-1}\left(\frac{8641\sqrt{759}}{1315347}\right)\right)+\frac{3}{10\sqrt{9241}}\sin \left(\mathrm{t}+{\tan }^{-1}\left(\frac{96}{5}\right)\right)$

Then, the frequency of the motion is:

$$\begin{gathered} \frac{{\gamma }_{\mathrm{r}}}{2\mathrm{\pi }}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}-\frac{{\mathrm{b}}^2}{2{\mathrm{m}}^2}} \\ =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{98}{2}-\frac{25}{8}} \\ =\frac{\sqrt{734}}{8\mathrm{\pi }} \end{gathered}$$

So, the frequency is $\frac{\sqrt{734}}{8\mathrm{\pi }}$.