In the following problems, take $\mathrm{g}=32\mathrm{ft}/{\sec }^2$for the U.S. Customary System and $\mathrm{g}=9.8\mathrm{m}/{\sec }^2$for the MKS system.

An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 40 N/m and the damping constant is 3 N/sec. At time t = 0, an external force $2\sin \left(2\mathrm{t}\right)\cos \left(2\mathrm{t}\right)$N is applied to the system. Determine the amplitude and frequency of the steady-state solution.

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 $\gamma$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 ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right),\omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$:

And the corresponding homogeneous equation is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$:

(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 of the system is $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

Given that, $\mathrm{m}=8$. And $\mathrm{F}=2\sin \left(2\mathrm{t}\right)\cos \left(2\mathrm{t}\right)=\sin \left(4\mathrm{t}\right)$.

So, ${\mathrm{F}}_0=1$and $\gamma \mathbf{=}4$. Since, $\mathrm{k}=40$and $\mathrm{b}=3$.

Then, the differential equation is $8\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+3\frac{\mathrm{dy}}{\mathrm{dt}}+40\mathrm{y}=\sin \left(4\mathrm{t}\right)$… (2)

Since one needs to find the steady state solution of the equation. That is all we are looking only for ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)$. Then,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\cos 4\mathrm{t}+{\mathrm{A}}_2\sin 4\mathrm{t}$… (3)

Now find the derivative of equation (3).

$$\begin{gathered} y_p^{\text{'}}\left(\mathrm{t}\right)=-4{\mathrm{A}}_1\sin 4\mathrm{t}+4{\mathrm{A}}_2\cos 4\mathrm{t} \\ {\mathrm{y}}_{\mathrm{p}}^{\text{'}\text{'}}\left(\mathrm{t}\right)=-16{\mathrm{A}}_1\cos 4\mathrm{t}-16{\mathrm{A}}_2\sin 4\mathrm{t} \end{gathered}$$

Substitute the values in equation (2).

$$\begin{gathered} 8\left(-16{\mathrm{A}}_1\cos 4\mathrm{t}-16{\mathrm{A}}_2\sin 4\mathrm{t}\right)+3\left(-4{\mathrm{A}}_1\sin 4\mathrm{t}+4{\mathrm{A}}_2\cos 4\mathrm{t}\right)+40\left({\mathrm{A}}_1\cos 4\mathrm{t}+{\mathrm{A}}_2\sin 4\mathrm{t}\right)=\sin \left(4\mathrm{t}\right) \\ \left(-128{\mathrm{A}}_1\cos 4\mathrm{t}-128{\mathrm{A}}_2\sin 4\mathrm{t}\right)+\left(-12{\mathrm{A}}_1\sin 4\mathrm{t}+12{\mathrm{A}}_2\cos 4\mathrm{t}\right)+\left(40{\mathrm{A}}_1\cos 4\mathrm{t}+40{\mathrm{A}}_2\sin 4\mathrm{t}\right)=\sin \left(4\mathrm{t}\right) \\ \left(-88{\mathrm{A}}_1+12{\mathrm{A}}_2\right)\cos \left(4\mathrm{t}\right)+\left(-12{\mathrm{A}}_1-88{\mathrm{A}}_2\right)\sin \left(4\mathrm{t}\right)=\sin \left(4\mathrm{t}\right) \end{gathered}$$

Now equalize the like terms.

$$\begin{gathered} -88{\mathrm{A}}_1+12{\mathrm{A}}_2=0 \\ {\mathrm{A}}_2=\frac{22}{3}{\mathrm{A}}_1 \\ -12{\mathrm{A}}_1-88{\mathrm{A}}_2=1 \\ -12{\mathrm{A}}_1-88\left(\frac{22}{3}{\mathrm{A}}_1\right)=1 \\ {\mathrm{A}}_1=-\frac{3}{1972} \\ {\mathrm{A}}_2=-\frac{11}{986} \end{gathered}$$

The solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{3}{1972}\cos \left(4\mathrm{t}\right)-\frac{11}{986}\sin \left(4\mathrm{t}\right)$.

From here, find the period.

$\mathrm{T}=\frac{2\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}$

Then, the frequency is $\mathrm{f}=\frac{1}{\mathrm{T}}=\frac{2}{\mathrm{\pi }}$.

Now, find the amplitude.

$$\begin{gathered} \mathrm{A}=\sqrt{A_1^2}+A_2^2 \\ =\sqrt{{\left(-\frac{3}{1972}\right)}^2+{\left(-\frac{11}{986}\right)}^2} \\ =\frac{\sqrt{493}}{1972} \\ \approx 0.011 \end{gathered}$$