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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
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1 Answer 1

up vote 1 down vote accepted

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?

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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
    
Aha!! Got it! They wanted the amplitude in the equation expressed in cm (why?? I assumed they'd want the SI unit). Anyways, the hint about the phase also helped. –  xbonez Apr 15 '12 at 22:28
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