Calculating Time for Simple Harmonic Motion - \pi

[rhodium]

\piHi everyone,

I hope you can help me out with the following question. I am mainly confused about the sign

Homework Statement

A block in SHM with T= 4.0 s and A=0.1 m. How long does it take the object to move from x=0 m to x= 0.06 m.

Homework Equations

x=Acos(wt + \phi)
w=2pi/T

where T is period, w is angular frequency, phi is phase angle and A is amplitude.

The Attempt at a Solution

Using eq2, i solved for w, which is \pi/2.
Then I set eq1 equal to 0. The value of phi is thus + or - \pi/2. Since the object is assumed to be moving to the right (as it would have to if it wants to go from 0 to 0.06 m),. then we take the negative phase angle. Then, back to eq1, I put

0.06=0.1cos((\pi/2)t - \pi/2)

NOW, this is were my problem is.
Apparently, there are then two possible answers,

either 0.927 = (\pi/2)t - \pi/2, which gives 1.59 s
or - 0.927 = (\pi/2)t - \pi/2. which gives 0.41 s

Firstly, I don't understand when are we supposed to have a + and - option. Secondly, I don't trust the answer given, which is 0.41 s. Please help.

 

[Doc Al]

Homework Equations

x=Acos(wt + \phi)
w=2pi/T

Note that you can also use:
x=Asin(wt + \phi)
Which would eliminate the annoying phase factor, since x = 0 at t = 0:
x=Asin(wt)

0.06=0.1cos((\pi/2)t - \pi/2)

NOW, this is were my problem is.
Apparently, there are then two possible answers,

either 0.927 = (\pi/2)t - \pi/2, which gives 1.59 s
or - 0.927 = (\pi/2)t - \pi/2. which gives 0.41 s

Realize that you want the first solution after t = 0. Since t = 0 corresponds to θ = -π/2 = -1.57 radians, θ = -0.927 radians is the first solution. The +0.927 solution corresponds to the block being at x = 0.06 but going to the left. (Since the block goes back and forth, there are an infinite number of times that it's at x = 0.06; you want the first time it reaches that point.)