Question 81 of 194

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement ($x$) starting from mean position to extreme position ($A$) is given by

ABCD

Show answer and solution▾

Answer: (C)

The formula to calculate the kinetic energy of a particle executing SHM is given by
$K=\dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{x}^{2}\right)...\left(1\right)$
Substitute $0$ for $x$ into equation (1) to obtain the kinetic energy at the mean position.
$\begin{matrix}K & = & \dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{0}^{2}\right) \\ & = & \dfrac{1}{2}m{\omega }^{2}{A}^{2}\end{matrix}$
Substitute $A$for $x$ into equation (1) to obtain the kinetic energy at the end points.
$\begin{matrix}K & = & \dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{A}^{2}\right) \\ & = & 0\end{matrix}$
Hence, the variation can be shown as follows-

JEE Main Physics practice question on Oscillations (Energy of Simple Harmonic Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.