91Ó°ÊÓ

The natural period of undamped force is$3\text{sec}$ .

And the period of damping force is $5\text{sec}$.

Angular frequency:

The frequency of a steadily recurring phenomenon expressed in radians per second.

In this caseknow that the differential equation that describe this system is

$m\frac{d^{2}x}{d{t}^2}+l\frac{dx}{dt}+kx=\mathrm{0.......}\left(1\right)$

which we can write it as

$\frac{d^{2}x}{d{t}^2}+\mathrm{η}\frac{dx}{dt}+{\mathrm{Ó\neg }}_{°}^{2}x=0$

here ,$\mathrm{η}=l/m=2b$ and${\mathrm{Ó\neg }}_{°}^2=k/m$ . For an undamming system the period was 3 seconds, so the angular frequency${\mathrm{Ó\neg }}_{°}$ is

$\begin{array}{c}{\mathrm{Ó\neg }}_{°}=\sqrt{\frac{k}{m}}\\ =\frac{2\mathrm{Ï€}}{T_{°}}\\ =\frac{2\mathrm{Ï€}}{3}\end{array}$

Here for the damping case, the period was $T=5\text{s}$,

The angular frequency for such system is

$\begin{array}{r}\begin{array}{l}{\mathrm{Ó\neg }}^{\text{'}2}={\mathrm{Ó\neg }}_{°}^2−\frac{l^2}{4m^2}\\ {\mathrm{Ó\neg }}^{\text{'}2}={\left(\frac{2\mathrm{Ï€}}{T}\right)}^2\\ {\mathrm{Ó\neg }}^{\text{'}2}=\frac{4{\mathrm{Ï€}}^2}{25}\end{array}\\ \end{array}$

Solve further

$\begin{array}{c}\frac{l}{m}=2\sqrt{{\mathrm{Ó\neg }}_{°}^2−{\mathrm{Ó\neg }}^{\text{'}2}}\\ =2\sqrt{\frac{4\mathrm{Ï€}}{9}−\frac{4{\mathrm{Ï€}}^2}{25}}\\ =\frac{16\mathrm{Ï€}}{15}\end{array}$

Thus, the equation of motion is

$\frac{d^{2}x}{d{t}^2}+\frac{16\mathrm{Ï€}}{15}\frac{dx}{dt}+\frac{4\mathrm{Ï€}}{9}x=0$

As notice $b<\mathrm{Ó\neg }$, , so this is an underdamped case, and the solution of this differential equation would be in the form of eq. (5.32), that is

$x=e^{−8\mathrm{π}t/15}(A\sin \frac{2\mathrm{π}}{5}t+B\cos \frac{2\mathrm{π}}{5}t)$