Question

The natural period of an undamped system is$3\text{sec}$, but with a damping force proportional to the velocity, the period becomes$5\text{sec}$. Find the differential equation of motion of the system and its solution.

Short answer

The equation of motion is,$\frac{d^{2}x}{d{t}^2}+\frac{16\pi }{15}\frac{dx}{dt}+\frac{4\pi }{9}x=0$ and the solution of it is$x=e^{-8\pi t/15}(A\sin \frac{2\pi }{5}t+B\cos \frac{2\pi }{5}t)$

Step-by-step solution

Given information

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

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

Angular frequency

Angular frequency:

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

Step 3:Solve for equation of motion

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

$\frac{d^{2}x}{d{t}^2}+\eta \frac{dx}{dt}+{\omega }_{°}^{2}x=0$

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

$\begin{array}{c}{\omega }_{°}=\sqrt{\frac{k}{m}}\\ =\frac{2\pi }{T_{°}}\\ =\frac{2\pi }{3}\end{array}$

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

Solution of the differential equation

The angular frequency for such system is

$\begin{array}{r}\begin{array}{l}{\omega }^{\text{'}2}={\omega }_{°}^2-\frac{l^2}{4m^2}\\ {\omega }^{\text{'}2}={\left(\frac{2\pi }{T}\right)}^2\\ {\omega }^{\text{'}2}=\frac{4{\pi }^2}{25}\end{array}\\ \end{array}$

Solve further

$\begin{array}{c}\frac{l}{m}=2\sqrt{{\omega }_{°}^2-{\omega }^{\text{'}2}}\\ =2\sqrt{\frac{4\pi }{9}-\frac{4{\pi }^2}{25}}\\ =\frac{16\pi }{15}\end{array}$

Equation of motion

Thus, the equation of motion is

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

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

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