In simple harmonic motion, is the phase, by definition, always measured using the sine function? I'm asking because a question came up that provided $\omega$ and the amplitude, and also specified the initial phase constant to be $\pi /4$.

Now, the answer says

$$A\sin(\omega t + \frac{\pi}{4})$$

but if you used the cosine function, it wouldn't be the same.


1 Answer 1


Physically, it does not matter whether we use $\sin()$ or $\cos()$. It's a matter of convention and you are free to choose either one.

Therefore, if somebody defined the amplitude, the frequency and the "initial phase" $\varphi_0 = 0$ of the oscillation, the function $x(t)$ is not (!) well defined. However, if we say, the initial phase is such that the harmonic oscillator starts with the maximal amplitude, we know $x(t) = A_0 \cos(\omega t)$. If instead we define the initial phase as the point with the minimal velocity (sign matters), we know that $v = - A_0 \omega \cos(\omega t)$. Hence, after integration we obtain $x(t) = A_0 \sin(\omega t)$.

  • $\begingroup$ Okay, so the question must be wrong. Thanks. $\endgroup$
    – harry
    Apr 3, 2020 at 2:53

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