Careers360 Logo
NITs Cutoff for BTech Civil Engineering - Check Opening & Closing Ranks

Energy In Simple Harmonic Motion - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Energy in SHM is considered one the most difficult concept.

  • 39 Questions around this concept.

Solve by difficulty

 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)

In a simple harmonic oscillator, at the mean position

A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x . Which of the following statements is true?

A particle of mass m executes simple harmonic motion with amplitude a  and frequency  \upsilon. The average kinetic energy during its motion from the  position of equilibrium to the end is :

A particle is executing simple harmonic motion with a time period T.  At time t=0, it is at its position of equilibrium.  The kinetic energy - time graph of the particle will look like :


The total energy of a particle, executing simple harmonic motion is:

Concepts Covered - 1

Energy in SHM

A particle executing S.H.M. possesses two types of energy: Potential energy and Kinetic energy

Potential energy-

  • This is an account of the displacement of the particle from its mean position.
  • Formula-

 As restoring force is given as F=-kx

        So U=-\int d w=-\int_{0}^{x} F d x=\int_{0}^{x} k x d x=\frac{1}{2} k x^{2}

      using \omega =\sqrt{\frac{k}{m}} or k= m\omega ^{2} 

      we get     U=\frac{1}{2} m \omega^{2} x^{2}

          For   x= A\sin \left ( wt \right )

         U=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t

  • Potential energy maximum and equal to total energy at extreme positions 

      i.e U_{\max }=\frac{1}{2} k A^{2}=\frac{1}{2} m \omega^{2} A^{2} \quad \text { when } x=\pm A ; \omega t=\pi / 2 ; \quad t=T / 4

  • Potential energy is minimum at mean position

       i.e U_{min}=0 \quad \text { when } x=0 ; \omega t=0 ; t=0

  • The average value of potential energy with respect to t  

                  Average\: o\! f \ U=\frac{\int U\; dt}{ \int dt}

                  \because U= \frac{1}{2}kx^{2}

                So U_{avg}= \frac{\int \frac{1}{2}m\omega ^{2}A^{2}\sin ^{2}\omega t}{\int dt}= \frac{\int \frac{1}{4} m \omega^{2} A^{2}(1-\cos 2 \omega t)dt}{dt}=\frac{1}{4}m\omega ^{2}A^{2} 

Kinetic energy-

  • This is because of the velocity of the particle.
  • Formula

            K=\frac{1}{2} m v^{2}

       or  using v=A\omega \cos\omega t  we get  K=\frac{1}{2} m A^{2} \omega^{2} \cos ^{2} \omega t

       And using v= w\sqrt{A^{2}-x^{2}}  and  k= m\omega ^{2} we get   K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) 

  • Kinetic energy is maximum at the mean position and equal to total energy at the mean position.

     i.e K_{\max }=\frac{1}{2} m \omega^{2} A^{2} \quad \text { when } x=0 ; t=0 ; \omega t=0

  • Kinetic energy is minimum at the extreme positions.

       i.e K_{\min }=0 \quad \text { when } y=A ; t=T / 4, \omega t=\pi / 2

  • The average value of kinetic energy with respect to t  

             K_{avg}=\frac{\int K\; dt}{ \int dt}

       K_{avg}= \frac{\int \frac{1}{2}m\omega ^{2}A^{2}\cos ^{2}\left ( \omega t \right )}{\int dt}= \frac{\int \frac{1}{4} m \omega^{2} A^{2}(1+\cos 2 \omega t)dt}{dt}=\frac{1}{4}m\omega ^{2}A^{2}

      So  K_{avg}= U_{avg}

Total energy-

  • Total mechanical energy = Kinetic energy + Potential energy  or E=K+U

E=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)+\frac{1}{2} m \omega^{2} x^{2}=\frac{1}{2} m \omega^{2} A^{2}

So Total energy does not depend on position(x)  i.e. it always remains constant in SHM.

  • Graph of Energy in S.H.M

At time t=0 sec, the position of the block is equal to the amplitude,





Study it with Videos

Energy in SHM

"Stay in the loop. Receive exam news, study resources, and expert advice!"


Reference Books

Energy in SHM

Physics Part II Textbook for Class XI

Page No. : 350

Line : 21

E-books & Sample Papers

Get Answer to all your questions

Back to top