Find the amplitude, period, frequency, and velocity amplitude for the motion of a particle whose distance from the origin is the given function.

$\mathit{s}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{t}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{t}$

Distance s from the origin is the given function$s=2\sin 3t\cos 3t$

Time Period (T): It is the time taken by the motion to repeat itself. So the unit of a time period is seconds.

Frequency (f) : It is defined as a number of times the motion is repeated in one second. The unit of frequency is Hz (Hertz).

Frequency is related to Time period as f= 1/T.

Following is the distance function:

$\begin{array}{rcl}s& =& 2\sin 3t\cos 3t\\ s& =& \sin (2\times 3t)\end{array}$

Since, $$\begin{gathered} 2\sin \theta \cos \theta =\sin 2\theta \\ s=\sin \left(6t\right) \end{gathered}$$

Amplitude is the maximum value of distance from the mean position.

So, amplitude is 1 .

The distance functions of the form.$A\sin \omega t$

So,$\omega =6$ .

Then, obtain values:

Period$=\frac{2\pi }{\omega }$

Period$=\frac{2\pi }{6}$

Period $=\frac{\pi }{3}$

The frequency is defined as the inverse of the period.

So,$\text{Frequency}=\frac{1}{\text{period}}$ .

Frequency$=\frac{3}{\pi }$

$$\begin{gathered} s=\sin 6t \\ \frac{ds}{dt}=6\cos 6t \end{gathered}$$

So, velocity amplitude is 6.