Chapter 11, Problem 17P

To determine

To Find:The distance from the equilibrium.

Answer to Problem 17P

The distance from the equilibrium at which the energy of Simple Harmonic Oscillators (SHO) is half KE and half PE is $\pm 0.707A$ .

Explanation of Solution

Introduction:

Energy of SHO is half kinetic and half potential, which indicates that the potential energy is half of the total energy.

  $E_p=\frac{1}{2}\times E_t$

Here.

  $E_p$ is the potential energy,

  $E_t$ is the total energy.

Total energy is the potential energy when the displacement have the value of amplitude. So it is expressed by the following way.

  $\begin{array}{l}{E}_p=\frac{1}{2}\times E_t\\ \frac{1}{2}\times k\times x^2=\frac{1}{2}\left(\frac{1}{2}\times k\times A^2\right)\\ x^2=\frac{A^2}{2}\\ x=\pm \frac{A}{\sqrt{2}}\\ x=\pm 0.707A\end{array}$

Here, $A$ is the amplitude or displacement at equilibrium condition.

Conclusion:

The distance from the equilibrium at which the energy of Simple Harmonic Oscillators (SHO) is half KE and half PE is $\pm 0.707A$ .