Name: Experiment: First Law of Simple Pendulum
Uploaded: 2019-07-19
Duration: 5 min 16 s
Description: Do you know how to measure the time period using a simple pendulum? The time period of a pendulum is independent of its amplitude when the amplitude is small...

Question

Derive the equation for the kinetic energy and potential energy of a body executing SHM and show that the total energy of a particle is proportional to the square of the amplitude

Accepted Answer

For a body executing simple harmonic motion, acceleration is given by, $a = - ω^{2} y$

where $ω, y$ are the angular velocity and displacement respectively. There fore the force on the body is
$F = ma = - {mω}^{2} y$

The work done in displacing the body by a small distance $dy$ is given by $force \times distance$
Or, work, $d w = - F d y \cos 180 °$ {because displacement and acceleration are in opposite direction so, $\cos 180 °$ ° taken}

The total work done
$W = \int dy = \int {mω}^{2} ydy = \frac{1}{2} {mω}^{2} y^{2} = \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt$

We know, Potential energy is work done stored in system.
so, potential energy $P = \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt$
Again, Kinetic energy, $K = \frac{1}{2} {mv}^{2}$ , here $v$ is velocity
we know, $v = Aω \cos ωt$
so, K.E = $K = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt$

Total energy $= K . E + P . E$
$= K = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt + \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt = \frac{1}{2} {mω}^{2} A^{2} (\cos^{2} ωt +^{2} \sin^{2} ωt) = \frac{1}{2} {mω}^{2} A^{2}$
which is a constant and proportional to the square of the amplitude $A$
Hence, total energy is always constant. A graphical representation is shown in the diagram

Derive the equation for the kinetic energy and potential energy of a body executing SHM and show that the total energy of a particle is proportional to the square of the amplitude

Important Questions on Simple Harmonic Motion

Learn and Prepare for any exam you want !