Question

For a mass-spring system that undergoes a simple harmonic motion of amplitude A, what are the positions (x =?) where the kinetic energy of the oscillator is 3 times the elastic potential energy (KE = 3PE,)

Answer 1

The kinetic energy in terms of the amplitude of oscillation is,

`K=(1)/(2)k(A^2-x^2)`

where k is the spring constant and x is the position,

The potential energy in terms of the amplitude of oscillation is,

`U=(1)/(2)kx^2`

The value of position when the kinetic energy of the oscillation is 3 times the potential energy is,

`\begin{gathered} K=3U \\ (1)/(2)k(A^2-x^2)=(3)/(2)kx^2 \\ k(A^2-x^2)=3kx^2 \\ kA^2-kx^2=3kx^2 \end{gathered}`

By simplifying,

`\begin{gathered} 3kx^2+kx^2=kA^2 \\ 4kx^2=kA^2 \\ x^2=(A^2)/(4) \\ x=\pm(A)/(2) \end{gathered}`

Thus, the position at which the kinetic energy becomes 3 times the potential energy is half of the amplitude.