# Simple Harmonic Motion of a Block on a Spring

1. Feb 7, 2010

### Neodudeman

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$\frac{1}{2}$$mv2 + $$\frac{1}{2}$$kx2 = Enet
x(t) = Acos($$\omega$$*t+$$\phi$$o)

3. 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? :/

2. Feb 7, 2010

### thebigstar25

I had a look at your solution .. maybe I cant 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 ..

3. Feb 7, 2010

### Neodudeman

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?