Assertion $(A)$: In $S.H.M$,kinetic and potential energy become equal when the distance is $1/\sqrt{2}$ times its amplitude. Reason $(R)$: The potential energy of a particle executing $S.H.M$ is periodic with time period being maximum at the extreme displacement.

  • A
    $A$ and $R$ are true. $R$ is the correct explanation of $A$.
  • B
    $A$ and $R$ are true. $R$ is not the correct explanation of $A$.
  • C
    $A$ is true,but $R$ is false.
  • D
    $A$ is false,but $R$ is true.

Explore More

Similar Questions

$A$ particle of mass $m$ oscillates with simple harmonic motion between points $x_1$ and $x_2$,the equilibrium position being $O$. Its potential energy is plotted. It will be as given below in the graph:

The displacement between the position of maximum potential energy and the position of maximum kinetic energy for a particle executing $S.H.M.$ is

$A$ particle executes $SHM$ of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

Position of a $3 \ kg$ mass moving along the $X$-axis is given by $x = 0.3 \cos (\omega t) \ m$. If $K(t)$ denotes the kinetic energy at time $t$,then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is

Show that for a particle in linear $SHM$,the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo