Harmonic motion: displacement

Question

On my textbook the displacement in a simple harmonic motion is given as
$$x = x(t)= A cos(omega t) tag 1$$
It is true that
$$x = x(t)= A cos(omega t)=A sin(pi/2-omega t ) tag 2$$
but why in some textbooks I find the law of simple harmonic motion as
$$bbox[yellow,5px,border:2px solid red]{x = x(t)= A sin(omega t)} qquad ? tag 3$$
Is there an advantage to use the $(3)$ instead of the $(1)$?

G. Smith · Accepted Answer

The general solution to
$$ddot x+omega^2 x=0$$
is
$$x(t)=Acosomega t+Bsinomega t.$$
(It’s a second-order differential equation, so there should be two integration constants.)
If you want $t=0$ to be when the pendulum is at top of its swing, choose the cosine. If you want $t=0$ to be when the pendulum is at the bottom, choose the sine. For other cases, you need both cosine and sine, or you need to use either a cosine or sine with a “phase angle”, which by a trigonometric identity is equivalent to a linear combination of a cosine and a sine.

Frobenius · Answer

Reference : My answer there Need help understanding an equation of motion for a pendulum.
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A choice for the general solution is the sinusoidal
begin{equation}
x_{bf s}left(tright)boldsymbol{=}alpha_{bf s}sin left(omega tboldsymbol{+}phi_{bf s}right) qquad alpha_{bf s},phi_{bf s} in mathbb{R}
tag{A-01}label{A-01}
end{equation}
The real constants $alpha_{bf s},phi_{bf s}$ are determined from the initial conditions of the harmonic oscillator.
This general expression is essentially identical to
begin{equation}
x(t)boldsymbol{=}Acosomega tboldsymbol{+}Bsinomega t  qquad A,Bin mathbb{R}
tag{A-02}label{A-02}
end{equation}
given by @G.Smith in his answer if we relate the two pairs of constants  $left(alpha,phiright)$ and $left(A,Bright)$ by
begin{equation}
alpha_{bf s}boldsymbol{=}sqrt{A^2boldsymbol{+}B^2}  qquadquad  phi_{bf s}boldsymbol{=}arcsindfrac{A}{sqrt{A^2boldsymbol{+}B^2}}boldsymbol{=}arccosdfrac{B}{sqrt{A^2boldsymbol{+}B^2}}
tag{A-03}label{A-03}
end{equation}
Equally well we could choice for the general solution the co-sinusoidal
begin{equation}
x_{bf c}left(tright)boldsymbol{=}alpha_{bf c}cos left(omega tboldsymbol{+}phi_{bf c}right) qquad alpha_{bf c},phi_{bf c} in mathbb{R}
tag{A-04}label{A-04}
end{equation}
The real constants $alpha_{bf c},phi_{bf c}$ are also determined from the initial conditions of the harmonic oscillator while their relations to the pair of constants  $left(A,Bright)$ of equation eqref{A-02} are given by
begin{equation}
alpha_{bf c}boldsymbol{=}sqrt{A^2boldsymbol{+}B^2}  qquadquad  phi_{bf c}boldsymbol{=}arccosdfrac{A}{sqrt{A^2boldsymbol{+}B^2}}boldsymbol{=}arcsindfrac{boldsymbol{-}B}{sqrt{A^2boldsymbol{+}B^2}}
tag{A-05}label{A-05}
end{equation}
$=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!=!$
An interpretation of the two alternative equivalent representations eqref{A-01} and eqref{A-04} of a simple harmonic oscillation is shown in the Figure below. Here a particle $:rm p:$ executes a  simple linear harmonic oscillation between the points $boldsymbol{-}R$ and $boldsymbol{+}R$ on the $xboldsymbol{-}$axis. Note that this motion could be considered as the projection on the $xboldsymbol{-}$axis of a plane uniform circular motion. If without loss of generality we have $omegaboldsymbol{>}0$ then the top circle represents an anticlockwise uniform circular motion corresponding to the co-sinusoidal choice eqref{A-04} while the bottom circle represents a clockwise uniform circular motion corresponding to the sinusoidal choice eqref{A-01}. The Figure shows the condition of the systems at time moment $tboldsymbol{=}0$.

Harmonic motion: displacement

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