#### Kintetic Energy of Particle Performing Linear S.H.M.:

- Consider a particle of mass ‘m’ which is performing linear S.H.M. of amplitude ‘a’ along straight line AB, with the centre O. Let the position of the particle at some instant be at C, at a distance x from O.

This is an expression for the kinetic energy of particle S.H.M.

Thus the kinetic energy of the particle performing linear S.H.M. and at a dis the ance of x_{1} from mean position is given by

#### Special cases:

**Case 1: Mean Position:**

Kinetic energy of particle performing S.H.M. is given by

For mean position x_{1} = 0

**Case 2: Extreme position:**

Kinetic energy of particle performing S.H.M. is given by

For mean position x_{1} = a

#### Potential Energy of Particle Performing Linear S.H.M.:

- Consider a particle of mass ‘m’ which is performing linear S.H.M. of amplitude ‘a’ along straight line AB, with the centre O. Let the position of the particle at some instant be at C, at a distance x from O.

Particle at C is acted upon by restoring force which is given by F = – mω²x

The negative sign indicates that force is restoring force.

- Let. External force F’ which is equal in magnitude and opposite to restoring force acts on the particle due to which the particle moves away from the mean position by small distance ‘dx’ as shown. Then

F’ = mω²x

Then the work done by force F’ is given by

dW = F’ . dx

dW = mω²x dx

The work done in moving the particle from position ‘O’ to ‘C’ can be calculated by integrating the above equation

This work will be stored in the particle as potential energy

This is an expression for the potential energy of particle performing S.H.M.

#### Special cases:

**Case 1: Mean Position:**

Potential energy of particle performing S.H.M. is given by

For mean position x_{1} = 0

∴ E_{P} = 0

**Case 2: Extreme position:**

Potential energy of particle performing S.H.M. is given by

For mean position x_{1} = a

#### Total Energy of Particle Performing Linear S.H.M.:

- The Kinetic energy of particle performing S.H.M. at a displacement of x
_{1}from mean position is given by

- The potential energy of particle performing S.H.M. at a displacement of x
_{1}from mean position is given by

- The total energy of particle performing S.H.M. at a displacement of x
_{1}from the mean position is given by

- Since for a given S.H.M., the mass of body m, angular speed ω and amplitude a are constant, Hence the total energy of a particle performing S.H.M. at C is constant. i.e. the total energy of a linear harmonic oscillator is conserved. It is same at all positions. The total energy of a linear harmonic oscillator is directly proportional to the square of its amplitude.

**Variation of Kinetic Energy and Potential Energy in S.H.M Graphically.**

**Relation Between the Total Energy of particle and Frequency of S.H.M.: **

**Relation Between the Total Energy and Period of S.H.M.: **

- The quantities in the bracket are constant. Therefore, the total energy of a linear harmonic oscillator is inversely proportional to the square of its period.

**Expressions for Potential Energy, Kinetic Energy and Total Energy of a Particle Performing S.H.M. in Terms of Force Constant:**

**Potential energy:**Potential energy of particle performing S.H.M. is given by

This is an expression for the potential energy of particle performing S.H.M. in terms of force constant.

**Kinetic energy:**Kinetic energy of particle performing S.H.M. is given by

This is an expression for Kinetic energy of particle performing S.H.M. in terms of force constant.

**Total energy:**Total energy of particle performing S.H.M. is given by

This is an expression for the total energy of particle performing S.H.M. in terms of force constant.

#### Numerical Problems:

#### Example – 1:

- A simple pendulum of length 1 m has a mass of 10 g and oscillates freely with an amplitude of 2 cm. Find its PE at the extreme point. g = 9.8 m/s².
**Solution:****Given:**Length of pendulum l = 1 m, Mass of bob = m = 10 g = 10 × 10^{-3}kg, amplitude = a = 2 cm = 2 × 10^{-2}m, g = 9.8 m/s².**To find:**Potential energy =? at x = a.- Ans: ( 1.96 x 10-5 J)