# What's wrong with this equation for harmonic oscillation?

The question:

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 1.70 cm, and the frequency is 1.10 Hz. Find an expression for the position of the particle as a function of time. (Use the following as necessary: t, and π.)

Using the equations:

$$x(t) = A \cos(\omega t + \Phi)$$ $$\omega = 2\pi f$$

I get A = 1.7cm or 0.017m, and $$\omega = 6.91$$

I know that t = 0, x = 0. Thus,

$$0 = 0.017 \cos(\Phi )$$

And therefore,

$$\Phi = \pi / 2$$

From all of this, it seems to me that the equation for position with respect to time should be:

$$x = 0.017 \cos(6.91t + \pi/2)$$

Am I doing something wrong, because the above is not getting checked as the right answer (it's an online homework)

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Is there some concept in here that you're not sure of, which you think might be responsible for the error? –  David Z Apr 15 '12 at 22:28
@DavidZaslavsky: I got it, finally. My phase was off by Pi, and the answer required the unit of Amplitude to be in cm. I was using m. –  xbonez Apr 15 '12 at 22:38
The cosine has more than one zero. And the text specifies that the particle goes to the right (I assume that the x axis also goes to the right). Now in which direction does the cosine go at $\pi/2$? And where's another zero?
I see your point, and so I tried $$x = 0.017 cos(6.91t + 3 \pi / 2)$$ as well, but it's still wrong! –  xbonez Apr 15 '12 at 22:24