Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

For SHM, acceleration , a = -ω²y F = ma = -mω²y now, work , W = F.dy cos180° { because displacement and acceleration are in opposite direction so, cos180° taken } W = ∫mω²y.dy = mω²y²/2 use standard form of SHM , y = Asin(ωt ± Ф) W = mω²A²/2 sin²(ωt ± Ф) We know, Potential energy is work done stored in system . so, P.E = W = mω²A²/2 sin²(ωt ± Ф) again, Kinetic energy , K.E = 1/2mv² , here v is velocity we know, v = ωAcos(ωt ± Ф) so, K.E = mω²A²/2cos²(ωt ± Ф) Total energy = K.E + P.E = mω²A²/2 cos²(ωt ± Ф) + mω²A²/2 sin²(ωt ± Ф) = mω²A²/2 [ cos²(ωt ± Ф) + sin²(ωt ± Ф) ] = mω²A²/2 [ ∵sin²α + cos²α = 1] = mω²A²/2 = constant

In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?

In case of SHM\\(\displaystyle KE = \frac{1}{2} m \omega^2 (a^2 - x^2)\\)and at \\(x =a/2\\)\\(\displaystyle KE = \frac{1}{2} m \omega^2 [a^2 - (a/2)^2]\\)\\(\displaystyle = \frac{3}{4} \left[ \frac{1}{2} m \omega^2 a^2 \right]\\)Total energy = \\(\displaystyle \frac{1}{2} m \omega^2 a^2\\)\\(\therefore\\) Fraction of total energy\\(\displaystyle = \frac{KE}{Total \,\, energy} = \frac{\displaystyle \frac{3}{4} \left[ \frac{1}{2}m \omega^2 a^2 \right] }{\displaystyle \frac{1}{2}m \omega a^2} = \frac{3}{4}\\)

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement (\\(amplitude =a\\)) is

\\(K.E.=P.E.\\ \Rightarrow \frac { 1 }{ 2 } m{ \omega }^{ 2 }({ a }^{ 2 }-{ x }^{ 2 })=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ x }^{ 2 }\\ \Rightarrow { a }^{ 2 }-{ x }^{ 2 }={ x }^{ 2 }\\ \Rightarrow 2{ x }^{ 2 }={ a }^{ 2 }\\ \Rightarrow x=a/\sqrt { 2 } \\)

Two springs of spring constant \\(1500\ N/m\\) and \\(3000\ N/m\\) respectively are stretched with the same force. They will have the potential energies in the ratio of

\\(KE=\dfrac{1}{2}*k*x^2\\)\\(k_1=3000N/m,k_2=1500N/m\\)\\(KE_1:KE_2=k_1:k_2=3000:1500=2:1\\) for same \\(x\\)

Starting from an origin, a body oscillates simple harmonically with a period of \\(2s\\). After what time will its kinetic energy be \\(75\%\\) of the total energy?

As per, \\( \displaystyle KE = \frac {75}{100} E = \frac {75}{100} \times \frac {1}{2} ma^2 \omega^2 \\)\\(\displaystyle \frac {1}{2} ma^2 \omega^2 \cos^2 \omega t = \frac {75}{100} \times \frac {1}{2} ma^2 \omega^2 \\)\\( \cos^2 \omega t = \dfrac {3}{4} \\)\\( \cos \omega t = \dfrac {\sqrt3}{2} = \cos \dfrac {\pi}{6} \\)or \\( \omega t = \dfrac {\pi}{6} \Rightarrow t = \dfrac {\pi}{6 \pi} = \dfrac {1}{6} s \\)

A block whose mass is \\(1kg\\) is fastened to a spring. The spring has a spring constant of \\(50{Nm}^{-1}\\). The block is pulled to a distance of \\(x=10cm\\) from its equilibrium position at \\(x=0\\) \\(cm\\) on a frictionless surface from rest at \\(t=0\\). The kinetic energy of the block when it is \\(5cm\\) away from the mean position is

Angular frequency of S.H.M\\(=\sqrt { \cfrac { k }{ { m } } } =5\sqrt { 2 } \\)Here, the body is displaced to maximum at \\(x=10cm=0.1m\\). So, it is the amplitude of the motion velocity of the body performing S.H.M at any position \\(x\\) then mean position is \\(v=\omega \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \\)Here \\(a=0.1m\\)\\(x=5cm=0.05m\\)\\(\Rightarrow v=5\sqrt { 2 } (\sqrt { 0.01-0.0005 } )\\)\\(v=0.61m/s\\)Kinetic energy at this point\\(=\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { 1 }{ 2 } \times 1\times { (0.61) }^{ 2 }\simeq 0.19J\\)

Two simple harmonic motions are represented by the equations \\({Y_1} = 10\;\sin \left( {3\pi t + \dfrac{\pi }{4}} \right)\\) and \\({Y_2} = 5\left( {\sin 3\pi t + \sqrt 3 \cos \;3\pi t} \right)\\)Their amplitudes are in the ratio of;

Given, \\(Y_1=10sin(3\pi t+\pi/4)\\) and \\(Y_2=5(sin3\pi t+\sqrt3 cos3\pi t)=5sin3\pi t+5\sqrt3 cos3\pi t\\)Amplitude of \\(Y_1\\) is \\(A_1=10\\) and amplitude of \\(Y_2\\) is \\(A_2=\sqrt{(5)^2+(5\sqrt 3)^2}=\sqrt{25+75}=10\\)Thus, \\(\dfrac{A_1}{A_2}=\dfrac{10}{10}=1:1\\)

Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M.

Acceleration of the particle , performing S.H.M is given by \\(\alpha = -\omega^2 y\\) where \\(\omega\\) is the angular velocity, and y is the displacement of particle. now, workdone by particle = \\(\vec F. \vec d y\\) as we know, acceleration and displacement are in opposite directions in case of S.H.M so, \\(W = -m\omega^2 y dy\\) where m is the mass of the particle. \\(W = -m\omega^2 \int ydy\\) \\(W = - \dfrac{1}{2} m \omega^2 y^2\\) so, potential energy = -W \\(=\dfrac{1}{2} m\omega^2 y^2\\) we know, \\(\omega = 2 \pi \eta\\) so, \\(P.E = 2 \pi^2 \eta^2 m y^2\\) ......(1) velocity of particle , \\(v = \omega Acos \omega t\\) or, \\(v = \omega \sqrt{A^2 - y^2}\\) so, kinetic energy of particle, \\(K.E = \dfrac{1}{2} mv^2\\) hence, \\(K.E = \dfrac{1}{2} m \omega^2 ( A^2 - y^2)\\) but \\(\omega = 2 \pi \eta\\) so, \\(K.E = 2 \pi^2 \eta^2 m ( A^2 - y^2)\\) ....(2) so, total mechanical energy = K.E + P.E \\(= 2 \pi^2 \eta^2 mA^2\\) ......(3)

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m,its time period would be

kinetic energy at mean position= total energy-potential energy at mean position\\(=9J-5J=4J\\)kinetic energy at mean position\\(=\dfrac { 1 }{ 2 } m{ v_{ max } }^{ 2 }\\)\\(\Rightarrow \dfrac { 1 }{ 2 } m{ v_{ max } }^{ 2 }=4J\\)\\(\Rightarrow { v_{ max } }=\sqrt { \dfrac { 4\times 2 }{ 2 } } m/s=2m/s\\)\\(\Rightarrow A\omega =A\dfrac { 2\pi }{ T } \\ \Rightarrow A\dfrac { 2\pi }{ T } =2m/s\\ \Rightarrow T=A\pi =0.01\pi =\dfrac { \pi }{ 100 } s\\)

Energy in SHM | Formulas, Definition, Examples

definition

Average kinetic energy and potential energy of a SHM

Total energy in SHM is given by,

E = \frac{1}{2} m ω^{2} A^{2}

where

A

is the amplitude and remains conserved.

E = K + U

K_{a v g} = U_{a v g} = \frac{E}{2} = \frac{1}{4} m ω^{2} A^{2}

Note:
Average kinetic energy can also be found using

K_{a v g} = \frac{1}{T} \int_{0}^{T} K d t

Average potential energy can also be found using

U_{a v g} = \frac{1}{T} \int_{0}^{T} U d t

formula

Use the relation between restoring force and potential energy

Restoring force is given by:

F = - \frac{d U}{d x}

It is often useful to find the equation of SHM.
Example:
A particle of mass

10

gm is placed in a potential field given by

V = (50 x^{2} + 100) J / k g

. Find the frequency of oscillation in cycle/sec.
Solution:
Potential energy

U = m V

\Rightarrow U = (50 x^{2} + 100) 1 0^{- 2}

F = - \frac{d U}{d x} = - (100 x) 1 0^{- 2}

\Rightarrow m ω^{2} x = - (100 \times 1 0^{- 2}) x

10 \times 1 0^{- 3} ω^{2} x = 100 \times 1 0^{- 2} x

\Rightarrow ω^{2} = 100, ω = 10

\Rightarrow f = \frac{ω}{2 π} = \frac{1 0}{2 π} = \frac{5}{π}

definition

Relation between restoring torque and potential energy

Restoring torque of a SHM can be found by:

τ = - \frac{d U}{d θ}

It is often useful in finding equation of SHM and helps in solving problems.

formula

Potential energy per unit length at a given point in a travellling sound wave

Potential energy per unit length

(w)

at a point is defined as:

w = \frac{p v}{c}

p is the sound pressure.
v is the particle velocity in the direction of propagation.
c is the speed of sound.

example

Kinetic energy per unit length at a given point in a travelling sound wave

Kinetic Energy of a traveling sound wave is defined as:

K . E . = \int \frac{ρ v ^{2}}{2} d V

Example. Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in figure. The speed of each pulse is 2 cm/s. After 2 seconds, what will be the total energy of the pulses.

Solution:
After two seconds, the two pulses would nullify each other. As the string now becomes straight, there would be no deformation of the string. In such a situation, there would be no potential energy.

formula

Total Potential Energy in one wavelength in a travelling sound wave

Total potential energy of a traveling sound wave is given by:

P . E . = \frac{1}{4} μ ω^{2} A^{2} λ

where,

μ =

Linear mass density,

ω =

Angular frequency,

A =

Amplitude of wave.

example

Derive and use kinetic energy per unit length at a given point in a travelling sound wave

Example:
The velocity of a sound wave in

v

and the wave energy density is

E,

then find the amount of energy transferred per unit area per second by the wave in a direction normal to the wave propagation.

Solution:
Energy transferred in normal direction to the wave propogation is also known as intensity

I = \frac{1}{2} ρ a^{2} ω^{2} v

E = ω^{2} a^{2} ρ

hence

I = E v

Energy in SHM

definition

Average kinetic energy and potential energy of a SHM

formula

Use the relation between restoring force and potential energy

definition

Relation between restoring torque and potential energy

formula

Potential energy per unit length at a given point in a travellling sound wave

example

Kinetic energy per unit length at a given point in a travelling sound wave

formula

Total Potential Energy in one wavelength in a travelling sound wave

example

Derive and use kinetic energy per unit length at a given point in a travelling sound wave

REVISE WITH CONCEPTS

Graphical Representation of Total Energy in SHM

Energy Equations of a Mass Attached to a Spring in SHM

Quick Summary With Stories

Energy in SHM

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

Draw graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.

In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement ( $a m p l i t u d e = a$ ) is

Two springs of spring constant $1500 N / m$ and $3000 N / m$ respectively are stretched with the same force. They will have the potential energies in the ratio of

Starting from an origin, a body oscillates simple harmonically with a period of $2 s$ . After what time will its kinetic energy be $75 %$ of the total energy?

Two simple harmonic motions are represented by the equations
$Y_{1} = 10 sin (3 π t + \frac{π}{4})$ and $Y_{2} = 5 (sin 3 π t + 3 cos 3 π t)$
Their amplitudes are in the ratio of;

Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M.

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m,its time period would be

definition

Average kinetic energy and potential energy of a SHM

formula

Use the relation between restoring force and potential energy

definition

Relation between restoring torque and potential energy

formula

Potential energy per unit length at a given point in a travellling sound wave

example

Kinetic energy per unit length at a given point in a travelling sound wave

formula

Total Potential Energy in one wavelength in a travelling sound wave

example

Derive and use kinetic energy per unit length at a given point in a travelling sound wave

REVISE WITH CONCEPTS

Graphical Representation of Total Energy in SHM

Energy Equations of a Mass Attached to a Spring in SHM

Quick Summary With Stories

Energy in SHM

Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion constant at any point on its path.

Draw graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.

In a SHM, when the displacement is one half the amplitude, what fraction of the total energy is kinetic?

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement (amplitude=a) is

Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have the potential energies in the ratio of

Starting from an origin, a body oscillates simple harmonically with a period of 2s. After what time will its kinetic energy be 75% of the total energy?

Two simple harmonic motions are represented by the equations Y1​=10sin(3πt+4π​) and Y2​=5(sin3πt+3​cos3πt) Their amplitudes are in the ratio of;

Obtain expressions for kinetic energy, potential energy and total energy of a particle performing linear S.H.M.

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m,its time period would be

The kinetic energy and potential energy of a particle executing SHM will be equal when displacement ( $a m p l i t u d e = a$ ) is

Two springs of spring constant $1500 N / m$ and $3000 N / m$ respectively are stretched with the same force. They will have the potential energies in the ratio of

Starting from an origin, a body oscillates simple harmonically with a period of $2 s$ . After what time will its kinetic energy be $75 %$ of the total energy?

Two simple harmonic motions are represented by the equations
$Y_{1} = 10 sin (3 π t + \frac{π}{4})$ and $Y_{2} = 5 (sin 3 π t + 3 cos 3 π t)$
Their amplitudes are in the ratio of;