Question

At what displacement of a SHO is the energy half kinetic and half potential?

Express your answer in terms of amplitude A.

Answer 1

Concepts and reason

The concept required to solve the given problems is kinetic energy, potential energy, and total energy of a simple harmonic oscillator.

Initially, investigate the expression for total mechanical energy of a simple harmonic oscillator. Then, investigate the expression for potential energy of a simple harmonic oscillator at a given displacement. Finally, equate half of total mechanical energy and potential energy to find the displacement.

Fundamentals

If there is no dissipative or frictional forces on a simple harmonic oscillator the kinetic and potential energies of the simple harmonic oscillator are alternately transformed each other but total mechanical energy is conserved. At equilibrium position the total energy of the simple harmonic oscillator is equal to kinetic energy because at equilibrium position the displacement is equal to zero and at extreme position the total energy is equal to potential energy because at extreme position speed is equal to zero.

Amplitude of a simple harmonic oscillator is defined as the maximum displacement from its equilibrium position.

The potential energy U of a simple harmonic oscillator at a given displacement x from the equilibrium position is given as follows:

$U = \frac{1}{2}k{x^2}$

Here, k is the spring constant.

The kinetic energy K of a simple harmonic oscillator at a given displacement is given as follows:

$K = \frac{1}{2}m{v^2}$

Here, m is the mass attached to the spring, and v is the speed of the mass at a given displacement.

The total mechanical energy E of a system is equal to the sum of potential energy U and kinetic energy K.

$E = U + K$

The total mechanical energy E of a system is equal to the sum of potential energy U and kinetic energy K.

$E = U + K$

Substitute $\frac{1}{2}k{x^2}$for U, and $\frac{1}{2}m{v^2}$for K, in the above equation and solve for E.

$E = \frac{1}{2}k{x^2} + \frac{1}{2}m{v^2}$

At extreme position, the displacement x is equal to amplitude A, and speed of oscillator is equal to zero.

Substitute A for x, and 0 for v in the above equation.

$\begin{array}{c}\\E = \frac{1}{2}k{A^2} + \frac{1}{2}m{\left( 0 \right)^2}\\\\ = \frac{1}{2}k{A^2}\\\end{array}$

The potential energy U of a simple harmonic oscillator at a given displacement x is,

$U = \frac{1}{2}k{x^2}$

Equate half of total mechanical energy and potential to find the displacement.

$\frac{E}{2} = U$

Substitute $\frac{1}{2}k{A^2}$for E, and $\frac{1}{2}k{x^2}$for U, in the above equation and solve for x.

$\begin{array}{c}\\\frac{1}{2}\left( {\frac{1}{2}k{A^2}} \right) = \left( {\frac{1}{2}k{x^2}} \right)\\\\\frac{{{A^2}}}{2} = {x^2}\\\\x = \frac{A}{{\sqrt 2 }}\\\end{array}$

Ans:

The displacement of SHO at which the energy half kinetic and half potential is $\frac{A}{{\sqrt 2 }}.$