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 $(1)$?


The general solution to

$$\ddot x+\omega^2 x=0$$


$$x(t)=A\cos\omega t+B\sin\omega t.$$

(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.

  • $\begingroup$ In the meantime thank you very much for your explanation that I have understood perfectly. I ask many simple questions because in the 3rd year of an high school, for Physics, they don't study trigonometry or goniometry. They have maybe the minimal basic knowledge of only sine and cosine. I understood everything and I sincerely confess that I did not know which of the two operators to choose when the pendulum is at the maximum of its oscillation (at the bottom) or at the lowest point. Please remember some textbook that concerns this fact? $\endgroup$ – Sebastiano Feb 7 at 18:40

Reference : My answer there Need help understanding an equation of motion for a pendulum.


A choice for the general solution is the sinusoidal \begin{equation} x_{\bf s}\left(t\right)\boldsymbol{=}\alpha_{\bf s}\sin \left(\omega t\boldsymbol{+}\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{=}A\cos\omega t\boldsymbol{+}B\sin\omega t \qquad A,B\in \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 $\left(\alpha,\phi\right)$ and $\left(A,B\right)$ by
\begin{equation} \alpha_{\bf s}\boldsymbol{=}\sqrt{A^2\boldsymbol{+}B^2} \qquad\quad \phi_{\bf s}\boldsymbol{=}\arcsin\dfrac{A}{\sqrt{A^2\boldsymbol{+}B^2}}\boldsymbol{=}\arccos\dfrac{B}{\sqrt{A^2\boldsymbol{+}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(t\right)\boldsymbol{=}\alpha_{\bf c}\cos \left(\omega t\boldsymbol{+}\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 $\left(A,B\right)$ of equation \eqref{A-02} are given by \begin{equation} \alpha_{\bf c}\boldsymbol{=}\sqrt{A^2\boldsymbol{+}B^2} \qquad\quad \phi_{\bf c}\boldsymbol{=}\arccos\dfrac{A}{\sqrt{A^2\boldsymbol{+}B^2}}\boldsymbol{=}\arcsin\dfrac{\boldsymbol{-}B}{\sqrt{A^2\boldsymbol{+}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 $x\boldsymbol{-}$axis. Note that this motion could be considered as the projection on the $x\boldsymbol{-}$axis of a plane uniform circular motion. If without loss of generality we have $\omega\boldsymbol{>}0$ 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 $t\boldsymbol{=}0$.

enter image description here

  • $\begingroup$ Yes I have understood, you have used the formula of the added angle that I explain in the fourth year when the students study goniometry in the detail. Always a sincere and affectionate thanks for the help you have given me over the years to you and to @G. Smith. My best regards. $\endgroup$ – Sebastiano Feb 7 at 18:43

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