Question

At a time when the displacement is $\dfrac{1}{2}$ the amplitude, what fraction of the total energy is kinetic energy, and what fraction is the potential energy in an S.H.M.

Answer 1

Hint: Simple harmonic motion is defined as the simplest form of periodic motion. An oscillatory motion is claimed to be simple harmonic when the displacement of the particle from the origin varies with time. We know that the velocity of a particle is the rate of change of displacement with time.
Formulae used:
The total energy of the system,
$T.E = \dfrac{1}{2}m{\omega ^2}{A^2}$
The kinetic energy
$K.E = \dfrac{1}{2}m{v^2}$
The velocity of the particle executing S.H.M
$\Rightarrow$ $v = \omega \sqrt {{A^2} - {x^2}} $
The potential energy,
$P.E = \dfrac{1}{2}m{\omega ^2}{x^2}$

Complete step by step solution:
Here let us consider the time to be $t$.
At a time $t$, the displacement of the particle will be half the amplitude of the particle.
i.e.
$x = \dfrac{1}{2}A$
The total energy of the particle undergoing S.H.M is given as,
$\Rightarrow$ $T.E = \dfrac{1}{2}m{\omega ^2}{A^2}$
where $m$ stands for the mass of the particle executing S.H.M, $\omega $ stands for the angular velocity, and $A$ stands for the amplitude.
The kinetic energy will be,
$\Rightarrow$ $K.E = \dfrac{1}{2}m{v^2}$
where $m$ stands for the mass of the particle executing S.H.M and $v$ stands for the velocity of the particle.
The velocity of the particle is given as,
$v = \omega \sqrt {{A^2} - {x^2}} $
where $\omega $ stands for the angular velocity, $A$ stands for the amplitude, and $x$ stands for the displacement of the particle.
Then the kinetic energy can be written as,
$K.E = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right)$
It is given that, the displacement $x = \dfrac{1}{2}A$
Substituting this value within the above equation, we get
$\Rightarrow$ $K.E = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - \dfrac{{{A^2}}}{4}} \right)$
This can be written as,
$K.E = \dfrac{1}{2}\dfrac{3}{4}m{\omega ^2}{A^2}$
We know that the total energy $T.E = \dfrac{1}{2}m{\omega ^2}{A^2}$
Therefore, the kinetic energy will be,
$K.E = \dfrac{3}{4}T.E$
The potential energy,
$\Rightarrow$ $P.E = \dfrac{1}{2}m{\omega ^2}{x^2}$
where $m$ stands for the mass of the particle executing S.H.M, $\omega $ stands for the angular velocity, and $x$ stands for the displacement of the particle.
Again putting, $x = \dfrac{1}{2}A$
$\Rightarrow$ $P.E = \dfrac{1}{2}m{\omega ^2}\dfrac{{{A^2}}}{4}$
This can be written as,
$P.E = \dfrac{1}{2}\dfrac{1}{4}m{\omega ^2}{A^2}$
In terms of total energy, this will be
$\Rightarrow$ $P.E = \dfrac{1}{4}T.E$
Therefore, at a time when the displacement is $\dfrac{1}{2}$ the amplitude, the potential energy will be $\dfrac{1}{4}$times the total energy, and the kinetic energy will be $\dfrac{3}{4}$ times the total energy.

Note:
In a simple harmonic oscillator, the total energy will be a constant. The total energy is only the kinetic energy at the mean position or only the potential energy at the extreme positions. The work done in restoring the force on the particle is stored as the potential energy.