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 Equations of a Mass Attached to a Spring in SHM

example

Write kinetic energy as a function of time in SHM

Kinetic energy as a function of time in SHM:

E =

\frac{m w ^{2} ( A ^{2} - A ^{2} s i n ^{2} ( w t + ϕ ) )}{2}

formula

Write potential energy as a function of time in SHM

Formula for potential energy as a function of time in SHM is:

P . E . = \frac{m w ^{2} x ^{2} s i n ^{2} ( w t + ϕ )}{2}

formula

Total energy as a function of time in SHM

Total energy as a function of time in SHM:

Total energy =

\frac{m w ^{2} ( A ^{2} )}{2}

(Independent of time)

example

Conservation of total mechanical energy to find amplitude

Example: Potential energy of a particle in SHM along

x -

axis is given by:

U = 10 + (x - 2)^{2}

.Here,

U

is in joule and

x

in meter. Total mechanical energy of the particle is

26 J .

Mass of the particle is

2 k g

Find the amplitude of oscillation.

Solution:

U_{0} =

minimum potential energy at mean position

(x = 2) = 10 J

At extreme position

U =

Total mechanical energy

= 26 J

= 10 + (x - 2)^{2}

∴

(x - 2) = \pm 4

Hence

x = 6 m

and

x = - 2 m

are the extreme positions.
Amplitude of oscillation

= 4 m

example

Write kinetic energy, potential energy and total energy of a mass attached to a spring in SHM

Kinetic Energy in SHM:

\frac{m w ^{2} ( A ^{2} - x ^{2} )}{2}

Potential Energy is :

\frac{m w ^{2} ( x ^{2} )}{2}

Total Energy is:

\frac{m w ^{2} ( A ^{2} )}{2}

Example: The potential energy of a simple pendulum in its resting position is

10

J and its mean kinetic energy is

5

J. What will be its total energy at any instant?

Solution:
The total energy of the system remains constant. Since it is given that P.E at rest is

10

J, the total energy must be

10

J as K.E at rest is 0. As total energy of the system is conserved in SHM.

example

Problem on kinetic energy, potential energy and total energy of a mass attached to a spring in SHM

Example: A mass

m

is attached to a spring of stiffness

k

executing SHM. It has amplitude

A

and velocity at the equilibrium position is

u

. Find the total energy of this spring mass system.

Solution:
At the extreme position of the spring it has only potential energy since velocity is zero:

\frac{k A ^{2}}{2}

At the equilibrium position it has no stretch in the spring.
Kinetic energy at this instant:

\frac{m u ^{2}}{2}

At any instant of time during the motion:
Total energy = KE + PE =

\frac{m u ^{2}}{2}

\frac{k A ^{2}}{2}

Energy Equations of a Mass Attached to a Spring in SHM

example

Write kinetic energy as a function of time in SHM

formula

Write potential energy as a function of time in SHM

formula

Total energy as a function of time in SHM

example

Conservation of total mechanical energy to find amplitude

example

Write kinetic energy, potential energy and total energy of a mass attached to a spring in SHM

example

Problem on kinetic energy, potential energy and total energy of a mass attached to a spring in SHM

REVISE WITH CONCEPTS

Energy in SHM

Graphical Representation of Total Energy in SHM

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