$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x+y)$ is

  • A
    $\sqrt{E}=\sqrt{E_1}-\sqrt{E_2}$
  • B
    $\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$
  • C
    $E=E_1-E_2$
  • D
    $E=E_1+E_2$

Explore More

Similar Questions

For a simple pendulum,a graph is plotted between its kinetic energy $(KE)$ and potential energy $(PE)$ against its displacement $d.$ Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

$A$ particle performs $S.H.M.$ from the mean position. Its amplitude is $A$ and total energy is $E$. At a particular instant,its kinetic energy is $\frac{3E}{4}$. The displacement of the particle at that instant is:

As a body performs $S.H.M.$,its potential energy $U$ varies with time $t$ as indicated in:

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t=\frac{T}{12}$,the ratio of the potential energy to kinetic energy of the particle is $\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right)$

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.

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