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

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where $E$ is the total energy)

The total energy of a particle executing $S.H.M.$ is proportional to

$A$ particle executing a simple harmonic motion of period $2 \ s$. When it is at its extreme displacement from its mean position,it receives an additional energy equal to what it had in its mean position. Due to this,in its subsequent motion,

Write the coordinates of the points of intersection of the graph of kinetic energy $(K)$ and potential energy $(U)$ for a simple harmonic oscillator.

$A$ mass of $1 \text{ kg}$ is executing $SHM$. Its displacement is given by $x = 6.0 \cos(100t + \pi/4) \text{ cm}$. What is the maximum kinetic energy (in $text{ J}$)?

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