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.

Determine the equation of motion for an undamped system at resonance governed by

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+9\mathrm{y}=2\cos 3\mathrm{t}; \\ \mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0. \end{gathered}$$

Sketch the solution.

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

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

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 given as; ${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right),\omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$. And the corresponding homogeneous equation is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

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,

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+9\mathrm{y}=2\cos 3\mathrm{t}; \\ \mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0. \end{gathered}$$

Then, m = 1, k = 9,and ${\mathrm{F}}_0=2and\gamma =3$.

Find the $\omega$value.

$$\begin{gathered} \omega =\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \\ =\sqrt{\frac{9}{1}} \\ =3 \end{gathered}$$

.

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

Find the derivative of y.

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{A}\omega \cos \left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\sin \omega \mathrm{t}+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tcos}\omega \mathrm{t}$

.

Given the initial conditions are $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0$.

Then,

$$\begin{gathered} \left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ \mathrm{y}\left(0\right)=\mathrm{Asin}\left(\omega \left(0\right)+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\left(0\right)\sin \omega \left(0\right) \\ 1=\mathrm{Asin}\left(\varphi \right) \end{gathered}$$

And

$$\begin{gathered} \left(\mathrm{t}\right)=\mathrm{A}\omega \cos \left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\sin \omega \mathrm{t}+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tcos}\omega \mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=\mathrm{A}\omega \cos \left(\omega \left(0\right)+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\sin \omega \left(0\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\left(0\right)\cos \omega \left(0\right) \\ 0=\mathrm{A}\omega \cos \left(\varphi \right) \\ 0=3\mathrm{Acos}\left(\varphi \right) \end{gathered}$$

So, A cannot be zero because $1=\mathrm{Asin}\left(\varphi \right)$.

Since $\cos \left(\varphi \right)=0$. Then,

$\varphi ={\cos }^{-1}\left(0\right)=\frac{\mathrm{\pi }}{2}+\mathrm{k\pi }$. Where k is an integer,

Case (1):

If k is even, k = 2l, then A becomes 1 and the solution can be written as:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)=\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}+\mathrm{k\pi }\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}+2\mathrm{l\pi }\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \end{gathered}$$

Case (2):

If k is odd, k = 2l + 1, then A becomes-1 and the solution can be written as:

$$\begin{gathered} \left(\mathrm{t}\right)=-\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}+2\mathrm{l\pi }+\mathrm{\pi }\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \end{gathered}$$

Since both cases are shown $\mathrm{y}\left(\mathrm{t}\right)=\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$. Then,

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)=\sin \left(\omega \mathrm{t}+\frac{\mathrm{\pi }}{2}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\cos \left(\omega \mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\cos \left(3\mathrm{t}\right)+\frac{2}{2\times 1\times 3}\mathrm{tsin}\left(3\mathrm{t}\right) \\ =\cos \left(3\mathrm{t}\right)+\frac{1}{3}\mathrm{tsin}\left(3\mathrm{t}\right) \end{gathered}$$

So, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\cos \left(3\mathrm{t}\right)+\frac{1}{3}\mathrm{tsin}\left(3\mathrm{t}\right)$

A sketch of the solution is shown below.