How should you start the system going at $t=0$ in order for the motion to be$\mathit{A}\mathit{c}\mathit{o}\mathit{s}\left(\omega t\right)$? How should you start the system going at in order for the motion to berole="math" localid="1654017955726" $\mathit{A}\mathit{s}\mathit{i}\mathit{n}\left(\omega t\right)$?

The time period,$t=0$ the motions are,

$x=A\cos \left(\omega t\right)$and$x=A\sin \left(\omega t\right).$

In basic words, a spring-mass system is one in which a block is hung or connected at the spring's free end. Any object executing a basic harmonic motion may generally be found using the spring-mass system.

Expression for analytical solution for spring-mass system is

$x=A\cos \left(\omega t\right)...........................\left(1\right)$

Where$A$ is the amplitude of the spring-mass system,$\omega$ is the angular frequency, and$t$ is the time period.

For the equation (1), motion is started given by,

$x=A\cos \left(\omega t\right)$

Now , the equation can be written as,

$x=A\cos \left(\omega \times 0\right)$

$$\begin{gathered} For\cos \left(0\right)=1 \\ x=A \end{gathered}$$

Then, $x=A_{max}$

Where A_max is the maximum amplitude.

Thus,$t=0$ , the system starts from the maximum amplitude.

Now for the motion

$x=A\sin \left(\omega t\right)$

Now $t=0$, the equation can be written as,

$$\begin{gathered} x-A\sin \left(\omega \times 0\right) \\ For\sin \left(0\right)=0 \\ x=0 \end{gathered}$$

Thus,$t=0$ , the system starts from its origin.