Question

A simple harmonic oscillator of amplitude A has a total energy E.

a) Determine the kinetic energy when the position is one-third the amplitude. (Use any variable or symbol stated above as necessary.) K =
b) Determine the potential energy when the position is one-third the amplitude. (Use any variable or symbol stated above as necessary.)

Answer 1

Answer:

a) $K=E-\frac{kA^2}{18}$

b) $U=\frac{kA^2}{18}$

Explanation:

The law of conservation of mechanical energy states that total mechanical energy remains constant during oscillation. Mechanical energy is defined as the sum of kinetic energy and potential energy:

$E=U+K\\E=\frac{kx^2}{2}+\frac{mv^2}{2}$

a) The position is one-third the amplitude. So, we have $x=\frac{1}{3}A$. Replacing and solving for K

$E=\frac{k(\frac{1}{3}A)^2}{2}+K\\E=\frac{kA^2}{18}+K\\K=E-\frac{kA^2}{18}$

b) The potential energy is defined as:

$U=\frac{kx^2}{2}$

Replacing:

$U=\frac{k(\frac{1}{3}A)^2}{2}\\U=\frac{kA^2}{18}$