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)
Best Answer
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?