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Numerical

Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potential energy

- at mean position and
- at extreme position.

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#### Solution

An oscillating particle possess both types of energies : Potential as well as kinetic. It possesses potential energy on account of its displacement from the equilibrium position.

Potential energy : Consider a particle of mass $m$ executing S.H.M. Let x be its displacement from the equilibrium position at any instant t. Since in S.H.M. the force F acting upon the particle is proportional to and opposite to the displacement x, we have

**`F = -kx`**

Where, the constant k gives the force per unit displacement $x$. The force can also be express in terms of potential energy (P.E) of the particle as :

`F = -(dw)/(dx)`

Thus, `(dw)/(dx) = kx`

On integrating, we get

`P.E = 1/2 kx^2 + C`

Where $C$ is a constant. If we assume the potential energy zero in the equilibrium position, i.e.,

If P.E = 0 at x = 0, then C = 0

Therefore, P.E = `1/2kx^2`

`"P"."E" = 1/2mw^2x^2`

Hence from here it is clear that the potential energy of a particle doing S.H.M. is directly proportional to the square of the displacement (P.E.`alpha x^2`)

(a) Potential energy at mean position : At mean position, velocity of a particle executing S.H.M. is maximum and displacement is minimum i.e., $0$

P.E = `1/2 mw^2x^2 = 0`

(b) Potential energy at extreme position : At extreme position, velocity of a particle executing S.H.M. is minimum and displacement is maximum, i.e., $±a$

P.E = `1/2 ka^2 = 1/2mw^2a^2`

Concept: K.E.(Kinetic Energy) and P.E.(Potential Energy) in S.H.M.

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