How to find phi in simple harmonic motion

[FlorenceC]

Given a position vs time graph of simple harmonic motion of an object and using the equation
x(t) = x_o = A sin (ωt +Φ), how am I supposed to find Φ?
I can easily find A and the period (T).
I also know T = 2 pi / ω, so I can find ω.

But I don't know x(t) and t. I tried finding random points in the graph and plugging them into the rearranged equation:
sin^-1(x(t)- x_o / (A)) - ωt = Φ
but I'm getting numbers that change depending on x(t) and t.

So how am I supposed to get Φ and should it not be constant?

 

[Simon Bridge]

Take a look at the x vs t graph at t=0.
You do know x(t) - you wrote it down on the second line of your post.
To get the specific version of x(t) for your graph, you will need to use the clues on the graph ... like you should be able to see what A is.

 

[FlorenceC]

Take a look at the x vs t graph at t=0.
You do know x(t) - you wrote it down on the second line of your post.
To get the specific version of x(t) for your graph, you will need to use the clues on the graph ... like you should be able to see what A is.

Yes but I don't know phi? How do I know x(t)?

 

[andrevdh]

You can use the value of x @ t = 0 seconds, say x_o, to determine the value of Φ.
Just watch out there are 2 points on an oscillation that have the same x value.
So when you solve for Φ using this x_o value you may not get the correct
angle.

 

[Simon Bridge]

Yes but I don't know phi? How do I know x(t)?

Take a look at the x vs t graph at t=0.
To get the specific version of x(t) for your graph, you will need to use the clues on the graph ... like you should be able to see what A is.

What is x(0)? What is A?

 

[lep11]

So you know T, A and ω. What is x when t=0?

 

[Simon Bridge]

So you know T, A and ω. What is x when t=0?

... That's probably about the most we can hope to do to help, without further feedback.
It kinda looks like using different values of t, knowing ω and A, gets different values for $\sin^{-1}\frac{x(t)}{A} -\omega t$ ... which suggests that there are easy mistakes to be made reading the graph.
Cannot tell though.