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# 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 $\dpi{100} x$ . Which of the following statements is true?

A particle of mass $\dpi{100} m$ executes simple harmonic motion with amplitude $\dpi{100} a$  and frequency  $\dpi{100} \upsilon$. 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

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## Books

### Reference Books

#### Energy in SHM

Physics Part II Textbook for Class XI

Page No. : 350

Line : 21