# Finding Phase Constant

1. Jun 16, 2009

### Liketothink

1. The problem statement, all variables and given/known data
An air-track glider attached to a spring oscillates with a period of 2.3 s. At t=0 s the glider is 3.09 cm left of the equilibrium position and moving to the right at 35.07 cm/s.
What is the phase constant? (in degrees)

2. Relevant equations

v0x= -wAsin(ro), x=Acos(ro)

3. The attempt at a solution
I divided v0x/x=wtan(ro). I solved for angular speed which was 2.73 rad/s. Now I solved for (ro) by taking tan inverse of (v/-wx). I received 73 as an answer and that's wrong. Would you please help me? Thank you.

2. Jun 16, 2009

### talaroue

Since you are to the left, you have to subtract 180. The anwser you came up with is to the right of the equilibrium point.

3. Jun 16, 2009

### Liketothink

I minus that by 180 so it's 107 but that's still wrong. What do you think is wrong?

4. Jun 16, 2009

### dx

Did you get

$$r_{0} = \tan^{-1} \frac{35.07}{3.09 \cdot 2.73}?$$

5. Jun 16, 2009

### Liketothink

yeah, I tried that as well which was 76.4751. Then I minus it from 180 so it's 103.525 but that's wrong.

6. Jun 16, 2009

### dx

Both 76.47 and 76.47 - 180 have the same tangent. You can see by drawing a pic that the one we want is (76.47 - 180).

7. Jun 16, 2009

### Liketothink

is it because it's negative and the distance is negative?

8. Jun 16, 2009

### Liketothink

Also, in this case, is angular speed the same as velocity? If I wanted to find the phase at t=0 s would I just need to use wt+ro, w being the velocity?

9. Jun 16, 2009

### dx

Yes, the cosine of 76.4 is positive, but the initial distance is negative.

10. Jun 16, 2009

### Liketothink

Oh sorry w=2.73. Sorry about that.

11. Jun 16, 2009

### dx

No, angular speed and velocity are different. The angular speed is constant, while the velocity is changing.

v = -Aωsin(ωt + a0)

12. Jun 16, 2009

### Liketothink

ok, so for t=.5s I get -102.115 but that's wrong. All I did is wt+constant(-103.48). But that's wrong for some reason. Does the constant change at that time?

13. Jun 16, 2009

### dx

14. Jun 16, 2009

### Liketothink

15. Jun 16, 2009

### dx

Do you get the right answer now? (I'm not sure what you're trying to calculate BTW, did you get the correct answer for the phase constant?)

16. Jun 16, 2009

### Liketothink

I got the correct answer for the phase constant but I'm trying the phase for t=.5s. Sorry for the confusion.

17. Jun 16, 2009

### dx

Ok, so at t = 0.5, you have ωt + a0 = 2.73(0.5) - 1.806. This is in radians. If you want to convert it to degrees, multiply by 360/2pi.

18. Jun 16, 2009

### Liketothink

The question has several parts and it's asking me for the phase at t=.5s. I used the same formula but it's wrong. Could you please let me know what's wrong? thank you.

19. Jun 16, 2009

### Liketothink

Ok thank you very much for all your help. I greatly appreciate it.