Question

A particle of mass m executes simple harmonic motion with amplitude ‘a’ and frequency ‘$v$’. The kinetic energy during its motion from the position of equilibrium to the end is:
A) ${\pi ^2}m{a^2}{v^2}$.
B) $\dfrac{1}{4} \cdot {\pi ^2}m{a^2}{v^2}$.
C) $4{\pi ^2}m{a^2}{v^2}$.
D) $2{\pi ^2}m{a^2}{v^2}$.

Answer 1

The simple harmonic motion is a motion in which the body is having to and fro motion between two extremes and also it passes through the mean position. In simple harmonic motion we know that the conversion of kinetic energy into potential energy as it passes through the mean position.

Formula used: The kinetic energy of the particle is given by $K.E = \dfrac{1}{2} \cdot m \cdot {v^2}$ and the motion can be given by $y = a\sin \omega t$.

Step by step solution:
The velocity of the particle is given by $v = \dfrac{{dy}}{{dt}}$.
$v = \dfrac{{dy}}{{dt}} = a\omega \cos \omega t$ and $\cos \omega t$.
$\cos \omega t = \sqrt {\dfrac{{{a^2} - {y^2}}}{{{a^2}}}}$
$v = \dfrac{{dy}}{{dt}} = a\omega \cos \omega t = \omega \sqrt {{a^2} - {y^2}}$
So, the kinetic energy is given by,
$K.E = \dfrac{{m{\omega ^2}\left( {{a^2} - {y^2}} \right)}}{2}$
So, the minimum kinetic energy of the particle is $K.E = 0$ when the particle is at extreme and the maximum kinetic energy of the particle is given by $K.E = \dfrac{{m{\omega ^2}{a^2}}}{2}$.
So the average kinetic energy of the particle is given by,
$K.{E_{avg}} = \dfrac{{K.{E_{\max .}} - K.{E_{\min .}}}}{2} \\\ K.{E_{avg}} = \dfrac{{m{\omega ^2}{a^2}}}{4} \\\$
Here the angular velocity is given as $\omega = 2\pi v$ where $v$ is the frequency of the particle. Replacing this value in the relation above.
$K.{E_{avg}} = \dfrac{{m{\omega ^2}{a^2}}}{4} \\\ K.{E_{avg}} = \dfrac{{m{{\left( {2\pi v} \right)}^2}{a^2}}}{4} \\\ K.{E_{avg}} = {\pi ^2}m{a^2}{v^2} \\\$

The correct answer for this problem is option A.

Note: Students should observe that the value of the kinetic energy will be different at different position of the simple harmonic position, at the two extreme positions the kinetic energy of the particle is zero as at that point the particle changes direction but at the mean position the kinetic energy the particle has maximum kinetic energy.