Why is the phase constant negative in x(t) = A cos(ωt + ϕ)?

[tyrostoken]

Homework Statement

A = 20 cm

Given the graph, what is the phase constant(in degrees)?
(I have already solved for amplitude and frequency)

Homework Equations

x(t) = A cos (wt + ϕ)

The Attempt at a Solution

x(t) = A cos (wt + ϕ)
Taking from t = 0, we find:
10 = 20cos(w*0 + ϕ)
10 = 20cos(ϕ)
1/2 = cos(ϕ)
cos^-1(1/2) = ϕ
ϕ = 60 degrees.

I tried submitting the answer and it said check signs, so I entered -60 degrees.
Why is this answer negative?

 

[gneill]

Hi tyrostoken, Welcome to Physics Forums!

The cosine trig function is many-to-one: Note that cos(Φ) = cos(-Φ). So when you solved for Φ the answer might have been either +60° or -60°. It's up to you to check which one fulfills the given conditions.

 

[haruspex]

To add to gneill's answer...
Imagine shifting the curve left until the value at t=0 is once again 0.5. This solves your equation, but the graph is visibly different. So you need another fact about the graph to fix the phase. What obvious difference is there in the graphs?