Simple Harmonic Motion of a Block on a Spring

[Neodudeman]

Homework Statement

There's a block attached to a spring on a frictionless surface that oscillates back and forth. (Assume no damping).
At t=0, the potential energy in the spring is 25% of the maximum potential energy.
Kinetic energy decreases with time at t=0, and at t=2, the kinetic energy becomes 0 for the first time.

Determine the frequency of this motion.

Homework Equations

\frac{1}{2}mv² + \frac{1}{2}kx² = E_net
x(t) = Acos(\omega*t+\phi_o)

The Attempt at a Solution

Ok. In order to get Frequency, we need the Period. To get the period, we solve for \omega by using the position function.
To use the position function, we must first find \phi.

So, according to the data, at t=0, the PE is 25% of the maximum potential energy. We know that the maximum potential energy is actually equal to \frac{1}{2}kA². Thus, 25% of the maximum potential energy is equal to \frac{1}{4}*\frac{1}{2}*kA².
Therefore, at t=0, \frac{1}{4}*\frac{1}{2}*kA²=\frac{1}{2}kx².

Solving for x, we get that x=\frac{1}{2}A.

Putting that into the position function,
@t=0
\frac{1}{2}A=Acos(\omega*0+\phi)
\frac{1}{2}=cos(\phi)
acos(\frac{1}{2}=\phi)

\phi=\pi/3

Now, solving for \omega and the period.

And this is where I have a problem...

x(t)=A*cos(\omega*t+\phi)
At t=2, the potential energy is max, meaning the kinetic energy is 0. Thus, the position x is equal to the amplitude A.

A=A*cos(\omega*2+\pi/3)
Divide by A, and \omega = 2\pi/T
1=cos(4\pi/T+\pi/3)
Acos(1) = 0
0 = 4\pi/T+\pi/3
Subtracting by \pi/3
-\pi/3 = 4\pi/T
-1/3=4/T
This gives us:
T=-12

Which, I'm 90% sure, we cannot have.
A negative period gives a negative frequency. Where did I mess up in this problem? :/

 

[thebigstar25]

I had a look at your solution .. maybe I can't be helpful .. but you know that cos(60)=cos(-60)=0.5 .. if you use the -60 instead of 60 (pi/3) you will end up with the right sign , since it is impossible (100%) to have a negative frequency or period they alwaye positive ..

 

[Neodudeman]

I had a look at your solution .. maybe I can't be helpful .. but you know that cos(60)=cos(-60)=0.5 .. if you use the -60 instead of 60 (pi/3) you will end up with the right sign , since it is impossible (100%) to have a negative frequency or period they alwaye positive ..

I think, expanding from this idea, it's possible to use, instead of
0=4\pi/3+\pi/3
you can do
2\pi=4\pi/3+\pi/3

This gives f=5/12

With a -\pi/3, it gives a f=1/12

Which is correct then? Hmm... Maybe someone can comment on the theories behind each of these approaches?