91Ó°ÊÓ

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

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

The angular frequency:

The amplitude of the steady-state solution to equation (1) depends on the angular frequency 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

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

The undamped system:

The system is governed by$\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \mathrm{γ}\mathrm{t}$ . 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

(21) ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\mathrm{Ó\neg }}\mathrm{tsin}\mathrm{Ó\neg }\mathrm{t}$. And the correct form is role="math" localid="1664018516305" ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcosÓ\neg t}+{\mathrm{A}}_2\mathrm{tsinÓ\neg t}$.

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, the weight mg = 8 (g = 32). Then,

$$\begin{gathered} \mathrm{m}=\frac{8}{32} \\ =\frac{1}{4} \end{gathered}$$

And$\mathrm{F}=2\cos 2\mathrm{t}$. So, ${\mathrm{F}}_0=2$ and$\mathrm{γ}=2$. Since, $\mathrm{k}=10$and$\mathrm{b}=1$.

Then, the differential equation is $\frac{1}{4}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\frac{\mathrm{dy}}{\mathrm{dt}}+10\mathrm{y}=2\cos \left(2\mathrm{t}\right)$

Since,

$$\begin{gathered} {\mathrm{b}}^2=1 \\ 4\mathrm{mk}=10 \end{gathered}$$

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

The homogeneous solution is given by

$$\begin{gathered} {\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\mathrm{Ï•}\right) \\ ={\mathrm{Ae}}^{-2\mathrm{t}}\sin \left(\frac{\sqrt{10-1}}{\frac{1}{2}}\mathrm{t}+\mathrm{Ï•}\right) \\ ={\mathrm{Ae}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+\mathrm{Ï•}\right) \end{gathered}$$

Where A and $\mathrm{Ï•}$ are constants.

Then, the particular solution is given by

$$\begin{gathered} {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\mathrm{γ}}^2\right)}^2+{\mathrm{b}}^2{\mathrm{γ}}^2}}\sin \left(\mathrm{γ}\mathrm{t}+\mathrm{θ}\right) \\ =\frac{2}{\sqrt{{\left(10-1\right)}^2+4}}\sin \left(2\mathrm{t}+\mathrm{θ}\right) \\ =\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+\mathrm{θ}\right) \end{gathered}$$

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+\mathrm{ϕ}\right)+\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+\mathrm{θ}\right)$ … (2)

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

Now find the derivative of equation (2).

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{A}\left(-2{\mathrm{e}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+\mathrm{ϕ}\right)+6{\mathrm{e}}^{-2\mathrm{t}}\cos \left(6\mathrm{t}+\mathrm{ϕ}\right)\right)+\frac{4}{\sqrt{85}}\cos \left(2\mathrm{t}+\mathrm{θ}\right)$

Now find the constants using the initial conditions.

$$\begin{gathered} \mathrm{y}\left(0\right)={\mathrm{Ae}}^{-2\left(0\right)}\sin \left(6\left(0\right)+\mathrm{ϕ}\right)+\frac{2}{\sqrt{85}}\sin \left(2\left(0\right)+\mathrm{θ}\right) \\ 0=\mathrm{Asin}\left(\mathrm{ϕ}\right)+\frac{2}{\sqrt{85}}\sin \left(\mathrm{θ}\right) \end{gathered}$$

So, $\mathrm{Asin}\left(\mathrm{ϕ}\right)+\frac{2}{\sqrt{85}}\sin \left(\mathrm{θ}\right)=0$…… (3)

$$\begin{gathered} \mathrm{y}\text{'}\left(0\right)=\mathrm{A}\left(-2{\mathrm{e}}^{-2\left(0\right)}\sin \left(6\left(0\right)+\mathrm{ϕ}\right)+6{\mathrm{e}}^{-2\left(0\right)}\cos \left(6\left(0\right)+\mathrm{ϕ}\right)\right)+\frac{4}{\sqrt{85}}\cos \left(2\left(0\right)+\mathrm{θ}\right) \\ 0=-2\mathrm{Asin}\left(\mathrm{ϕ}\right)+6\mathrm{Acos}\left(\mathrm{ϕ}\right)+\frac{4}{\sqrt{85}}\cos \left(\mathrm{θ}\right) \end{gathered}$$

So, $-2\mathrm{Asin}\left(\mathrm{ϕ}\right)+6\mathrm{Acos}\left(\mathrm{ϕ}\right)+\frac{4}{\sqrt{85}}\cos \left(\mathrm{θ}\right)=0$........(4)

Now solve the equation (3) and (4)

$$\begin{gathered} \frac{\mathrm{Asin}\left(\mathrm{Ï•}\right)}{-2\mathrm{Asin}\left(\mathrm{Ï•}\right)+6\mathrm{Acos}\left(\mathrm{Ï•}\right)}=\frac{\frac{2}{\sqrt{85}}\sin \left(\mathrm{θ}\right)}{\frac{4}{\sqrt{85}}\cos \left(\mathrm{θ}\right)} \\ \frac{\tan \mathrm{Ï•}}{2\tan \mathrm{Ï•}-6}=\frac{1}{2}\mathrm{³Ù²¹²Ôθ} \\ \tan \mathrm{Ï•}=\frac{9}{4}\left(2\tan \mathrm{Ï•}-6\right) \\ =\frac{27}{11} \\ \mathrm{Ï•}={\tan }^{-1}\left(\frac{27}{11}\right)+\mathrm{°ìÏ€} \end{gathered}$$

Then, find the value of A.

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

$$\begin{gathered} \mathrm{A}=\frac{\frac{2}{\sqrt{85}}\sin \left({\tan }^{-1}\left(\frac{9}{2}\right)\right)}{-\sin \left({\tan }^{-1}\left(\frac{27}{11}\right)\right)} \\ =-\frac{2\sqrt{34}}{51} \end{gathered}$$

So, the solution is$\mathrm{y}\left(\mathrm{t}\right)=-\frac{2\sqrt{34}}{51}{\mathrm{e}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+{\tan }^{-1}\left(\frac{27}{11}\right)\right)+\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+{\tan }^{-1}\left(\frac{9}{2}\right)\right)$ .

Then, the frequency of the motion is

$$\begin{gathered} \frac{{\mathrm{γ}}_{\mathrm{r}}}{2\mathrm{π}}=\frac{1}{2\mathrm{π}}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}-\frac{{\mathrm{b}}^2}{2{\mathrm{m}}^2}} \\ =\frac{1}{2\mathrm{π}}\sqrt{40-8} \\ =\frac{4\sqrt{2}}{2\mathrm{π}} \\ =\frac{2\sqrt{2}}{\mathrm{π}} \end{gathered}$$

So, the frequency is$\frac{2\sqrt{2}}{\mathrm{Ï€}}$ .