91Ó°ÊÓ

The undamped system:

The system is governed by $m\frac{d^{2}y}{d{t}^2}+ky=F_0\cos \mathrm{γ}t$. And the homogenous solution of it is $y_h\left(t\right)=A\sin \left(\mathrm{Ó\neg } t+\mathrm{Ï•}\right), \mathrm{Ó\neg }\text{:}=\sqrt{\frac{k}{m}}$.

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

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

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 $A\left(\mathrm{γ}\right)=F_{0}M\left(\mathrm{γ}\right)$, where

(13)role="math" localid="1663950251816" $M\left(\mathrm{γ}\right)=\frac{1}{\sqrt{{\left(k-m{\mathrm{γ}}^2\right)}^2+b^2{\mathrm{γ}}^2}}      ......\left(1\right)$

To derive the formula of$y_P\left(t\right)$

Referring to equation (20): one gets,

$y_p\left(t\right)=A_1 t\cos \left(\mathrm{Ó\neg } t\right)+A_2 t\sin  \left(\mathrm{Ó\neg } t\right)$. And the differential equation isrole="math" localid="1663950351011" $m\frac{d^{2}y}{d{t}^2}+ky=F_{0}cos\left(\mathrm{Ó\neg }t\right)      ......\left(2\right)$

Then, substitute equation (20) with equation (2).

$m\left(A_1 t\cos \left(\mathrm{Ó\neg } t\right)+A_2 t\sin \left(\mathrm{Ó\neg } t\right)\right)\text{'}\text{'}+k\left(A_1 t\sin \left(\mathrm{Ó\neg } t\right)+A_2 t\sin \left(\mathrm{Ó\neg } t\right)\right)=F_0\cos \left(\mathrm{Ó\neg } t\right)      ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Find the first and second-order derivatives of y.

$$\begin{gathered} y{\text{'}}_p\left(t\right)=A_1\cos \left(\mathrm{Ó\neg } t\right)-\mathrm{Ó\neg } A_1 t\sin \left(\mathrm{Ó\neg } t\right)+A_2\sin \left(\mathrm{Ó\neg } t\right)+\mathrm{Ó\neg } A_2 t\cos \left(\mathrm{Ó\neg } t\right) \\ y\text{'}{\text{'}}_p\left(t\right)=\mathrm{Ó\neg } A_1\sin \left(\mathrm{Ó\neg } t\right)-{\mathrm{Ó\neg }}^2 A_1 t\cos \left(\mathrm{Ó\neg } t\right)+\mathrm{Ó\neg } A_2\cos \left(\mathrm{Ó\neg } t\right)-{\mathrm{Ó\neg }}^2 A_2 t\sin \left(\mathrm{Ó\neg } t\right) \end{gathered}$$

Substitute the second derivative in equation (3).

$$\begin{gathered} m\left(- \mathrm{Ó\neg } A_1\sin \left(\mathrm{Ó\neg } t\right)-{\mathrm{Ó\neg }}^2 A_1 t\cos \left(\mathrm{Ó\neg } t\right)+\mathrm{Ó\neg } A_2\cos \left(\mathrm{Ó\neg } t\right)-{\mathrm{Ó\neg }}^2{A}_2 t\sin \left(\mathrm{Ó\neg } t\right)\right)+k\left(A_1 t\cos \left(\mathrm{Ó\neg } t\right)+A_2 t\sin \left(\mathrm{Ó\neg } t\right)\right)=F_0\cos \left(\mathrm{Ó\neg } t\right) \\ -m \mathrm{Ó\neg } A_1\sin \left(\mathrm{Ó\neg } t\right)-m {\mathrm{Ó\neg }}^2{A}_1 t\cos \left(\mathrm{Ó\neg } t\right)+m \mathrm{Ó\neg } A_2\cos \left(\mathrm{Ó\neg } t\right)-m {\mathrm{Ó\neg }}^2{A}_2 t\sin \left(\mathrm{Ó\neg } t\right)+k{A}_1 t\cos \left(\mathrm{Ó\neg } t\right)+k{A}_2 t\sin \left(\mathrm{Ó\neg } t\right)=F_0\cos \left(\mathrm{Ó\neg } t\right) \end{gathered}$$

Now we equate the coefficients on the left and right sides. To get,

$$\begin{gathered} 2m{A}_2 \mathrm{Ó\neg }=F_0 \\ -2m{A}_1 \mathrm{Ó\neg }=0 \end{gathered}$$

Then,

$$\begin{gathered} A_2=\frac{F_0}{2m\mathrm{Ó\neg }} \\ {A}_1=0 \end{gathered}$$

So, the solution is$y_P\left(t\right)=\frac{F_0}{2\mathrm{Ï–}}ts\mathrm{in}\left(\mathrm{Ï–}t\right)$