# 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.

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 $\pi$).

Also remember that the "displacement equation" you refer to could just as well be: $$x(t) = A \sin{\left(\omega t + \phi\right)}$$ The phase angle $\phi$ is a measure of when you decide to start your stopwatch, i.e. when you set $t=0$. For an easy way to remember this, just remember the graphs of $\sin{t}$ and $\cos{t}$ -- the sine function has the value $0$ at $t=0$, while the cosine function has a value $1$ at $t=0$. Thus, if you start your clock when the oscillating body is at the equilibrium position, you should use the sine, and if you start your clock when the oscillating body is at the maximum amplitude position, you should use the cosine.