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

Derive the formula for given in (21).

The undamped system:

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

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

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

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

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

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

Referring to equation (20): one gets,

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

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

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

Find the first and second-order derivatives of y.

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

Substitute the second derivative in equation (3).

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

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

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

Then,

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

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