Total energy for a simple harmonic oscillator:

$\overline{){{\mathbf{E}}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}{\mathbf{=}}{\mathbf{U}}{\mathbf{+}}{\mathbf{K}}}$

Potential energy U = (1/2)kA^{2}

Kinetic energy, K = (1/2) mv^{2}

If the kinetic energy, K = (3/4)E, we have:

$\begin{array}{rcl}\mathbf{E}& \mathbf{=}& \mathbf{U}\mathbf{+}\frac{\mathbf{3}}{\mathbf{4}}\mathbf{E}\\ \begin{array}{rc}\mathbf{E}\mathbf{-}& \frac{\mathbf{3}}{\mathbf{4}}\mathbf{E}\end{array}& \mathbf{=}& \mathbf{U}\\ \mathbf{U}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}}\mathbf{E}\end{array}$

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.

What is the system's potential energy when its kinetic energy is equal to 3/4 E?

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