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Next: 15-36. Up: Energy I: Week 1 Previous: 15-28.


The period $T=1/f$, while $f=\omega/2\pi$, and $\omega^2=k/m$. Thus we need to find $k$. We know $a_{max}$, which can be utilized in a few different ways. The most staightforward is the use the force equation:

\begin{displaymath}F=ma=-kx \to F_{max}=ma_{max}=kA.\end{displaymath}

Thus we have


Doing all of the back substitutions yields $T=3.1\;{\textrm ms}$.

Next, we want to find the maximum velocity. Since velocity is the derivative of position we have

\begin{displaymath}v(t)=\frac{dx}{dt}=-A\omega\sin(\omega t + \varphi).\end{displaymath}

So, for our maximum value we have


We know from 28 that we can use the maximum speed or displacement to calculate the total energy. Choosing displacement, we have


Dan Cross 2006-10-18