91Ó°ÊÓ

The angular frequency:

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

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

The undamped system:

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

So, the general solution of the system is $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{Ó\neg }\mathrm{t}+\mathrm{Ï•}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t}$.

Given that,

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{y}=5\mathrm{cost}; \\ \mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=1. \end{gathered}$$

Then, m = 1, k = 1, and ${\mathrm{F}}_0=5and\mathrm{γ}=1$.

Find the$\mathrm{Ó\neg }$ value.

$$\begin{gathered} \backslash b\mathrm{Ó\neg }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \\ =\sqrt{\frac{1}{1}} \\ =1 \end{gathered}$$

.

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

Find the derivative of y.

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

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

Then,

$$\begin{gathered} \left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{Ó\neg }\mathrm{t}+\mathrm{Ï•}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \\ \mathrm{y}\left(0\right)=\mathrm{Asin}\left(\mathrm{Ó\neg }\left(0\right)+\mathrm{Ï•}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\left(0\right)\sin \mathrm{Ó\neg }\left(0\right) \\ 0=\mathrm{Asin}\left(\mathrm{Ï•}\right) \end{gathered}$$

And

$$\begin{gathered} \left(\mathrm{t}\right)=\mathrm{A}\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }\mathrm{t}+\mathrm{Ï•}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\sin \mathrm{Ó\neg }\mathrm{t}+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tcos}\mathrm{Ó\neg }\mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=\mathrm{A}\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }\left(0\right)+\mathrm{Ï•}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\sin \mathrm{Ó\neg }\left(0\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\left(0\right)\cos \mathrm{Ó\neg }\left(0\right) \\ 0=\mathrm{A}\mathrm{Ó\neg }\cos \left(\mathrm{Ï•}\right) \\ 1=\mathrm{Acos}\left(\mathrm{Ï•}\right) \end{gathered}$$

So, A cannot be zero because $0=\mathrm{Asin}\left(\mathrm{Ï•}\right)$.

Since $\sin \left(\mathrm{Ï•}\right)=0$. Then, $\mathrm{Ï•}={\sin }^{-1}\left(0\right)=\mathrm{°ìÏ€}$, Where k belongs to an integer.

Case (1):

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

$$\begin{gathered} \left(\mathrm{t}\right)=\sin \left(\mathrm{Ó\neg }\mathrm{t}+\mathrm{°ìÏ€}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \\ =\sin \left(\mathrm{Ó\neg }\mathrm{t}+2\mathrm{\pm ôÏ€}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \\ =\sin \left(\mathrm{Ó\neg }\mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \end{gathered}$$

Case (2):

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

$$\begin{gathered} \left(\mathrm{t}\right)=-\sin \left(\mathrm{Ó\neg }\mathrm{t}+2\mathrm{\pm ôÏ€}+\mathrm{Ï€}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \\ =\sin \left(\mathrm{Ó\neg }\mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \end{gathered}$$

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

$$\begin{gathered} \left(\mathrm{t}\right)=\sin \left(\mathrm{Ó\neg }\mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t} \\ =\sin \left(\mathrm{t}\right)+\frac{5}{2×1×1}\mathrm{tsin}\left(\mathrm{t}\right) \\ =\sin \left(\mathrm{t}\right)+\frac{5}{2}\mathrm{tsin}\left(\mathrm{t}\right) \end{gathered}$$

So, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\sin \left(\mathrm{t}\right)+\frac{5}{2}\mathrm{tsin}\left(\mathrm{t}\right)$.

A sketch of the solution is shown below.