# Question #ba86e

Mar 22, 2017

See below

#### Explanation:

I think it's more interesting if we try to work it from the paradigm SHM for a particle of mass $m$, ie without digging out equations that from memory might lead us straight there.

So, for displacement, we have the generalised idea of SHM:

$x \left(t\right) = A \cos \left(\omega t + \psi\right)$

Kinetic Energy: $T = \frac{1}{2} m {\left(\dot{x}\right)}^{2}$

Velocity: $\dot{x} = - \omega A \sin \left(\omega t + \psi\right)$

$\implies T \left(t\right) = \frac{1}{2} m {\omega}^{2} {A}^{2} {\sin}^{2} \left(\omega t + \psi\right)$

And:
${\sin}^{2} \theta \in \left[0 , 1\right] \implies {T}_{\max} = \frac{1}{2} m {\omega}^{2} {A}^{2}$

SHM is conservative and is in essence a system where energy oscillates between potential and kinetic energy. So at certain points in time, the energy is all either Kinetic or Potential Energy.

If follows for Potential Energy $U \left(t\right)$ that:

${U}_{\max} = \frac{1}{2} m {\omega}^{2} {A}^{2}$

In terms of Total Energy ${E}_{T} \left(t\right)$, we also can see that:

${E}_{T} \left(t\right) = T \left(t\right) + U \left(t\right) = c o n s t \implies {E}_{T} = \frac{1}{2} m {\omega}^{2} {A}^{2}$

Now at $x = \frac{A}{2}$, and if at that time, $t = \tau$ , then:

$\frac{A}{2} = A \cos \left(\omega \tau + \psi\right) \implies \cos \left(\omega \tau + \psi\right) = \frac{1}{2}$

$T \left(\tau\right) = \frac{1}{2} m {\omega}^{2} {A}^{2} {\sin}^{2} \left(\omega \tau + \psi\right)$

$= \frac{1}{2} m {\omega}^{2} {A}^{2} \left(1 - {\cos}^{2} \left(\omega \tau + \psi\right)\right)$

$= \frac{3}{8} m {\omega}^{2} {A}^{2}$

So

$\frac{T \left(\tau\right)}{{E}_{T} \left(\tau\right)} = \frac{\frac{3}{8} m {\omega}^{2} {A}^{2}}{\frac{1}{2} m {\omega}^{2} {A}^{2}}$

$= \textcolor{b l u e}{\frac{3}{4}}$

And:

$\frac{U \left(\tau\right)}{{E}_{T} \left(\tau\right)} = 1 - \frac{3}{4} = \textcolor{b l u e}{\frac{1}{4}}$