Question

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

$\mathit{s}\mathbf{=}\mathbf{5}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{t}\mathbf{-}\mathit{\pi }\mathbf{)}$

Short answer

The velocity amplitude of motion of a particle is 5.

Step-by-step solution

Given data

Distance from the origin the given function is$s=5\sin (t-\pi )$

Concept of Periodic motion formula

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.

Calculation of the amplitude of motion of a particle

Following is the distance function:

$s=5\sin (t-\pi )$

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

So, amplitude = 5 .

Calculation of the period of motion of a particle

The distance functions of the form$A\sin (\omega t-\varphi )$.

So, $\omega =1$.

Then, obtain values:

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

Period $=\frac{2\pi }{1}$

Period $=2\pi$

Calculation of the velocity amplitude of motion of a particle

The frequency is defined as the inverse of the period.

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

Frequency $\frac{1}{2\pi }$

$\begin{array}{rcl}s& =& 5\sin (t-\pi )\\ \frac{ds}{dt}& =& 5\cos (t-\pi )\end{array}$

So, velocity amplitude is 5.