# Simple Harmonic Motion - seemingly easy yet

1. Dec 1, 2005

The function x = (2.0 m) cos[(2pi rad/s)t + pi/2 rad] gives the simple harmonic motion of a body. Find the following values at t = 4.0 s.
(a) the displacement: ____m
Correct me if I am wrong, but to get x all I have to do is just plug 4.0 s in for t in that equation mentioned above right? I set my calculator to radians and did what I just said and get 6E^-13, but for some reason webassign (website where I answer the problem at) says I'm wrong. What gives?

Last edited: Dec 1, 2005
2. Dec 1, 2005

### Staff: Mentor

That's all there is to it.
You made a mistake. What angle (in radians) are you taking the cosine of?

3. Dec 1, 2005

Oh I'm sorry, I forgot to include the pi in the equation.

4. Dec 1, 2005

### Staff: Mentor

So I assume you corrected your mistake?

5. Dec 1, 2005

Oh I haven't, I just forgot to include it when I posted the problem. I still do not understand why I am not getting the right answer when all I have to do is plug in 4.0 for T.

6. Dec 1, 2005

### mezarashi

I think if you just take a look at the trigonometric function itself, it should be no surprise to you that any integer value of t will result in the cosine function returning 0.

7. Dec 2, 2005

### Staff: Mentor

mezarashi explained it, but your real mistake is using a calculator to solve this. If you just looked at the equation, you'd see that plugging in T = 4 sec gives you an angle of pi/2 radians. What's the cosine of pi/2 radians (or 90 degrees)? (Note that 8 1/2 pi radians is equivalent to pi/2 radians since the trig functions are periodic over 2 pi radians.)

Your calculator answer was "correct": 6E^-13 is pretty close to zero! (Round it off.)

Last edited: Dec 2, 2005
8. Dec 2, 2005