# Harmonic oscillation displacement

1. Nov 3, 2016

### Vitani11

1. The problem statement, all variables and given/known data
...when she pulls the ball down 2.5cm from equilibrium and releases it from rest, it oscillates at 5.5 Hz. What is displacement y as functions of t?

2. Relevant equations
Y= Acos(omega t+phi)

3. The attempt at a solution
I'm almost certain I should instead be using sin to characterize the displacement or add in a phase angle of pi to the cosine function since its released from bottom. Is this a correct assumption?

2. Nov 3, 2016

### Staff: Mentor

You can use either sin or cos to describe harmonic motion. Whether or not you need to incorporate a phase constant depends upon your choice of coordinate system (is the down direction considered positive or negative?) and where you want to place your initial position. As long as the resulting math describes your scenario as interpreted though your choices, everything's good.

3. Nov 3, 2016

### Vitani11

Yes I know this but I don't have any confidence in my ability to do physics so I need a clarification that this is not just cosine without a phase change, but sine or cosine with a phase change

4. Nov 3, 2016

### Staff: Mentor

You should be able to try them all and check them by plugging in a few values to see if what you've got matches your scenario. That's how you make the knowledge stick.

Ultimately whether or not you'll need a phase constant depends on how you choose your coordinate system. In this case if you chose "down" to be positive then a simple cosine will do, or a sine with a phase shift of $\pi/2$. If "up" is positive (so that the initial displacement is taken to be negative) then a cosine or a sine, both with phase shifts, will work.

Sketch out a few wavelengths of a sine and a cosine function. Mark the major angles along the axis. You should be able to see how sine and cosine are the same basic form with just a phase shift to distinguish them. You can use a sketch like this to choose a function and phase shift to suit your purpose.

5. Nov 3, 2016

### rude man

No, because at t=0 your y displacement value is 2.5cm. If you use the sine (without a pi/2 addition in the argument) you'd get y=0 at t=0 which is wrong.

6. Nov 4, 2016

### Vitani11

Agreed I finished this yesterday and turned it in - thank you though.

7. Nov 4, 2016

### BvU

So do you want to keep us in suspense until you get it back marked, or reassure us that you submitted 'displacement from equilibrium $= -2.5 \cos (2\pi\;5.5\; t)$ ' ?
(Oops...)

8. Nov 4, 2016

### Vitani11

Lol I did exactly that except I defined down to be positive so it was −2.5cos(2π5.5t) without the sign.

9. Nov 4, 2016

### Vitani11

I won't have it back marked for a week