homework and exercises - Which trigonometric ratio should be used to describe simple harmonic motion as a function of time?

An object is undergoing simple harmonic motion with period 1.2s and amplitude 0.6m. At $t=0$, the object is at $x=0$. How far is the object from the equilibrium position when $t=0.480$s?

I used the displacement equation :

$$x(t)=A\cos(\omega t+\phi)$$

and also found out what the angular frequency $\omega$ is (5.2rad/s). Then I found that $\phi$ is 0. I plugged my results in the equation and when I looked at the solution they used the following equation:

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

How was I supposed to know to use this equation instead of the cosine equation that is written on my equation sheet.

Answer

The initial phase $\phi$ should be $-\pi/2$, or in other words, the solution should be $A\sin(\omega t)$ because $\cos(\theta-\pi/2)=\sin(\theta)$. How did you find that $\phi$ is zero? You're supposed to use the fact that $x$ is zero when $t$ is zero. Which is what let's you set $\phi$ to $-\pi/2$ (modulo $\pi$).