# What is the phase constant?

1. Nov 8, 2006

### ubbaken

What is the phase constant? Use a cosine function to describe the simple harmonic motion.

http://capa.physics.mcmaster.ca/figures/kn/Graph14/kn-pic1416_new.png [Broken]

t1=40.0 s and A=20.0 cm

I'm really lost on how to get this done. Using x=Acos(ωt+φ); my approach has been to find when cos()=1 or 0 and then setting the phase equal to either 0 or pi/2 respectively.

So far, I have ω=pi/10 rad and I know pi<φ<3pi/2.

I'm stuck past this though (didn't get very far)...I tried it assuming the graph was accurate in scale and that it was 225deg (5pi/4 rad) but it was incorrect. I've also tried -10=20cos((pi/10)0+φ) --> φ=120deg; then adding a fraction of the period to get it in the proper phase, getting 240deg or 210deg (tried all, incorrect).

I am sure there is a more mathematical way to do it, that I am just missing or that I am starting, but not finishing correctly. Any help would be appreciated. Thanks!

Last edited by a moderator: May 2, 2017
2. Nov 8, 2006

### Mindscrape

Okay, well I would suggest using the less mathematical way, and use the graph they have given you because it will help more than fiddeling around with numbers. At time=0 they show you that the amplitude is -1/2 of its max value. At time=0 a cosine function without a phase constant will be at the maximum amplitude. For a moment forget about the time and amplitude they give you. Phase constants, if we use the conventional definition of a positive phase constant, shift the function to the left.

You can formulate an equation to figure out the argument that belongs in the cosine function, and find the phase constant from there.

Hopefully that helps without giving you a direct formula.

Last edited: Nov 8, 2006
3. Nov 8, 2006

### ubbaken

tried that, though I didn't word it that well there it seems.

with that logic, I work through to get: -1/2=cos(φ); which gives me 120deg - which isn't in the correct phase, so continuing along the curve until 240deg gives the right phase, but the answer is coming up as incorrect.

4. Nov 8, 2006

### Mindscrape

I am not really sure where we went wrong then. At t=0 x=-(1/2)A, so $$-\frac{1}{2} A = Acos(\omega*(0) + \phi)$$. It should be at 2(pi)/3. That doesn't work? Anyone else understand where we wrong?

Are you using an online thing that wants you to use radians?

Last edited: Nov 8, 2006
5. Nov 8, 2006

### ubbaken

but it can't be 2pi/3 rads, cause that isn't the phase, it's the next one equal to -1/2. But I've tried them both. It accepts deg or rad answers.

6. Nov 8, 2006

### Mindscrape

Why can't it be 2pi/3? We want when the first -1/2 comes about, which is at the first phase shift. The other one, at 4pi/3, will go past the lower amplitude, and then go back up to -1/2.

Are you supposed to compose the whole waveform, solving for the angular frequency and all?

7. Nov 27, 2008

### Tzar_MacEng

Heya, I have this exact same problem. I played around a bit. Recall "The quantity (wt + φ) is called the phase, and is measured in radians. The cosine function traces out one complete cycle when the phase changes by 2p radians. The phase is not a physical angle!" So, by that and the graph, all you do is make your answer for the question negative. I hope that helps. I had a different question and I kept getting 2.09 rad as the answer, I made it negative and it worked. On the 7th try. Darn CAPA.

Sorry for the formatting(Equation and grammar wise), I was googling around for an answer and I found this, then I solved the question, figured I'd post where I went wrong.

8. Jun 11, 2011

### valsess

I had to solve the same question for physics class. Solution: 4 pi - 2 pi /3 = 10.5 rad (using the formula wt - initial phase).