Question

At what distance from the mean position of SHM of
amplitude r the energy is half kinetic energy and half potential energy​

Answer 1

Answer:

the energy is half kinetic energy and half potential energy​ at total amplitude/√2

Answer 2

Given:

A particle is undergoing Simple Harmonic Motion (SHM) having an amplitude $r$.

To Find:

Distance from the mean at which the kinetic energy is equal to the potential energy.

Solution:

We know that a particle undergoing SHM has total energy $E$ given by

$E=\frac{1}{2} kr^2$ where $k$ is the force constant and $r$ is the amplitude of oscillation.

The total energy at any point i.e. the sum of the kinetic energy and the potential energy always remains constant.

When kinetic energy and potential energy become equal, the potential energy becomes half of the total energy

     $\frac{1}{2} kx^2=\frac{1}{2}* \frac{1}{2} kr^2$

⇒ $x^2=\frac{1}{2} r^2$

⇒ $x=\frac{r}{\sqrt{2} }$

Hence, at a distance  $\frac{r}{\sqrt{2} }$ from the mean position, the kinetic energy, and the potential energy become equal.