Hey students we know that, the particle performing linear SHM vibrates about its mean position, so that it possess mass and velocity. Due to this the particle performing linear SHM is said to have the potential energy as well as kinetic energy. The total energy of particle then can be obtained by taking sum of its kinetic energy and potential energy.

**Let’s obtain the energy for particle in linear SHM……..!**

**Kinetic energy of particle in linear SHM:-**

Consider a particle of mass m performing linear SHM with angular velocity ω. Let x be the displacement of particle from its mean position, as shown in fig below

Let the particle at any point P at distance ‘x’ from O, then KE of particle is,

KE = 1/2 mv^{2} ……..(1)

The instantaneous velocity of particle performing SHM is given as,

Then equation (1) becomes,

KE = 1/2 mω^{2} (a^{2} – x^{2}) …………………..(2)

But we know that the displacement in SHM is given as,

∴ x = a sin (ωt + α)

On substituting in above equation we get,

∴ KE = 1/2 mω^{2} (a^{2} – (a sin (ωt + α)^{2})

∴ KE = 1/2 mω^{2} (a^{2} – a^{2} sin^{2} (ωt + α))

∴ KE = 1/2 mω^{2} a^{2} (1- sin^{2} (ωt + α))

∴ KE = 1/2 mω^{2} a^{2} cos^{2} (ωt + α) …………………(A)

Equation (A) gives the kinetic energy of particle performing linear SHM.

**Potential energy of particle in linear SHM:-**

When particle is at point P, it is acted upon by restoring force directed towards mean position. The particle is displaced through small distance ‘dx’ against the restoring force. The small work done is,

∴ dW = -Fdx …….(3)

The negative sign indicates that work done is against the restoring force.

∴ dW = k x dx ……(4)

The total work done can be obtain by integrating the equation (2) from *0 to x.*

This work is stored in the form potential energy.

∴ PE=1/2 mω^{2} x^{2} …………….(5)

But we know that the displacement in SHM is given as,

∴ x = a sin (ωt + α)

On substituting in above equation we get,

∴PE = 1/2 mω^{2} a^{2} sin^{2} (ωt + α) ……….(B)

**Total energy of particle in linear SHM:-**

The total energy of particle performing SHM is sum of its KE and PE.

∴ Total energy = Kinetic energy + Potential energy

On substituting the values of KE and PE we get from equation (2) and (5) we get,

As, ω = 2πf then the above equation becomes

∴ TE = 1/2 m (2πf)^{2} a^{2}

∴ TE = 1/2 m4 π^{2} f^{2} a^{2}

The above equation gives the total energy of particle performing linear SHM.

**Some important points……….!**

1) total energy is directly proportional to

- Square of amplitude (a
^{2}) - Square of frequency of particle (n
^{2})

2) Total energy of particle in SHM is nothing but kinetic energy at mean position, i.e.

At mean position, x=0

∴ PE = 0 and KE = 1/2 mω^{2} a^{2}

∴TE = 0 + 1/2 mω^{2} a^{2}

∴TE = 1/2 mω^{2} a^{2}

3) Total energy of particle in SHM is nothing but potential energy at extreme position,

At extreme position, x = a

∴ KE = 0 and PE = 1/2 mω^{2} a^{2}

∴TE = 1/2 mω^{2} a^{2} + 0

∴ TE = 1/2 mω^{2} a^{2}

**Let’s solve some numerical to understand the concept with more details……….!**

Ex:-1) A particle performing linear SHM with amplitude 4 cm. Find the ratio of kinetic energy to potential energy when the particle at distance 3 cm.

Solution:

Here, A= 4 cm, x= 3 cm

Ex:2) The force constant of material of spring is 12 N/m. Find the Kinetic energy, potential energy and total energy of particle performing linear SHM with amplitude of 15 cm and at distance of 5 cm from mean position.

Solution:-

Here, K = 12 N/m, A = 15 cm =15 × 10 – 2 m, x = 5 cm = 5 × 10 – 2 m

We have, KE = 1/2 k (a^{2} -x^{2})

∴ KE = 1/2 × 12 × (15^{2} – 5^{2}) × 10^{-4}

∴KE = 6 × (225 – 25) × 10^{-4}

∴KE = 6 × 200 × 10^{-4}

∴KE = 1200 × 10^{-4} J

Now,

∴PE = 1/2 kx^{2}

∴PE = 1/2 × 12 × 5^{2} × 10^{-4}

∴PE = 6 × 25 × 10^{-4 }

∴PE =150 × 10^{-4} J

The total energy is sum of KE and PE

∴TE = KE + PE

∴TE = 1200 × 10^{-4} + 150 × 10^{-4}

∴TE = 1350 × 10^{-4} J