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

a)
`K=E-(kA^2)/(18)`

b)
`U=(kA^2)/(18)`

Step-by-step 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=(kx^2)/(2)+(mv^2)/(2)`

a) The position is one-third the amplitude. So, we have
`x=(1)/(3)A`. Replacing and solving for K

`E=(k((1)/(3)A)^2)/(2)+K\\E=(kA^2)/(18)+K\\K=E-(kA^2)/(18)`

b) The potential energy is defined as:

`U=(kx^2)/(2)`

Replacing:

`U=(k((1)/(3)A)^2)/(2)\\U=(kA^2)/(18)`