## Simple Harmonic Motion: Displacement after s seconds

The question is:

A 56.0 kg bungy-jumper hangs suspended from her bungy-cord, at rest. She is displaced from this position by 15.0 m downward, and then released. She bounces up and down, with a period of 5.800 s. Assume the woman undergoes simple harmonic motion, described by
y(t) = X cos(2 π t / T + φ)

Where is the woman after 43.20 s of bouncing? (enter a negative value if she is below her rest position).

The answer is 14.2m.

I've been using x=Xcos(2*pi*t/T) --> 15cos(2pi43.2/5.8) = 10.27m. This equation was in my book. What did I do wrong? Am I supposed to use y(t) = X cos(2 π t / T + φ? In that case, I don't know what φ is...
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 Mentor Blog Entries: 1 Are you using your calculator properly? (Radians, not degrees.)
 Oooh, thanks. I did not know you had to use radians. So do you have to always have to use radians when dealing with simple harmonic motion?

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## Simple Harmonic Motion: Displacement after s seconds

 Quote by mexqwerty Oooh, thanks. I did not know you had to use radians. So do you have to always have to use radians when dealing with simple harmonic motion?
Generally, yes. But you can always convert from one to the other.

Note what's going on in the equation y(t) = X cos(2 π t / T). That ratio t/T tells you what fraction of a period you are dealing with. The 2π tells you that you are dealing with radians, since one complete period is 2π radians.
 Oh, that make sense. Thanks, that was really helpful. =)