# About energy in Simple Harmonic Motion

The problem is this;

Given the amplitude A, find the a value of x for which the kinetic energy represents half of the total energy of the system.

I have so far tried to relate A to a maximum height when considering an ideal pendulum but the relationship is hard to derive. The instructor recommended using a spring mass system for the problem.

Any ideas?

• Try to writing down the following expressions: the mechanical energy in terms of the amplitude, the mechanical energy in terms of the kinetic and potential energies, the potential energy in therms of the position and the kinetic energy as one half of the mechanical energy. It won't take you more than one line of math to solve this. – Diracology Jun 3 '16 at 22:09

## 1 Answer

For simple harmonic motion kinetic energy kinetic can be written as $^1/_2mv^2$ which evaluates to $^1/_2m\omega^2 (A^2-x^2)$ where $\omega=\sqrt{^k/_m}$, $A$ is amplitude $k$ is spring constant and $x$ is displacement from equilibrium position. So at $x=0$ kinetic energy will be equal to total energy which will be $^1/_2mA^2\omega^2$. Now according to your condition $KE=^1/_2TE$

$\implies ^1/_2m\omega^2(A^2-x^2)=\frac{1}{2}^1/_2m\omega^2A^2$

$\implies \frac{A^2}{2}=A^2-x^2$

$\implies 2x^2=A^2$ This gives that at $x=\frac{A}{\sqrt{2}}$ kinetic energy would be half of total energy.