(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex]\frac{1}{2}[/tex]mv^{2}+ [tex]\frac{1}{2}[/tex]kx^{2}= E_{net}

x(t) = Acos([tex]\omega[/tex]*t+[tex]\phi[/tex]_{o})

3. The attempt at a solution

Ok. In order to get Frequency, we need the Period. To get the period, we solve for [tex]\omega[/tex] by using the position function.

To use the position function, we must first find [tex]\phi[/tex].

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 [tex]\frac{1}{2}[/tex]kA^{2}. Thus, 25% of the maximum potential energy is equal to [tex]\frac{1}{4}[/tex]*[tex]\frac{1}{2}[/tex]*kA^{2}.

Therefore, at t=0, [tex]\frac{1}{4}[/tex]*[tex]\frac{1}{2}[/tex]*kA^{2}=[tex]\frac{1}{2}[/tex]kx^{2}.

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

Putting that into the position function,

@t=0

[tex]\frac{1}{2}[/tex]A=Acos([tex]\omega[/tex]*0+[tex]\phi[/tex])

[tex]\frac{1}{2}[/tex]=cos([tex]\phi[/tex])

acos([tex]\frac{1}{2}[/tex]=[tex]\phi[/tex])

[tex]\phi[/tex]=[tex]\pi[/tex]/3

Now, solving for [tex]\omega[/tex] and the period.

And this is where I have a problem...

x(t)=A*cos([tex]\omega[/tex]*t+[tex]\phi[/tex])

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([tex]\omega[/tex]*2+[tex]\pi[/tex]/3)

Divide by A, and [tex]\omega[/tex] = 2[tex]\pi[/tex]/T

1=cos(4[tex]\pi[/tex]/T+[tex]\pi[/tex]/3)

Acos(1) = 0

0 = 4[tex]\pi[/tex]/T+[tex]\pi[/tex]/3

Subtracting by [tex]\pi[/tex]/3

-[tex]\pi[/tex]/3 = 4[tex]\pi[/tex]/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? :/

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# Simple Harmonic Motion of a Block on a Spring

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