Obtan an expression for potential energy of a particle performing S.H.M. What is the value of potential energy at (i) Mean position, and (ii) Extreme position

#### Solution

Potential Energy :

→ consider a particle of mass ‘m’ performing S.H.M along AB about mean position ‘O’.

→ OB = OA = a.

∴ When particle of mass ‘m’ performs S.H.M. is at distance x from its mean position.

→ Let ‘F = k x’ be the restoring force acting on the particle where k → force constant.

→ Let the particle is displaced by infinitesimal small distance 'dx'.

∴ Workdone,

dw = - fdx= -(-kx) dx= kxdx.

∴ total workdone is ,

`∫ dw= ∫ _(x=0)^x kxdx =k ∫ _(x=0)^x xdx`

`W = 1/2 kx^2= 1/2 mω^2x^2` `[∵ ω^2 = k/m]`

∴ this workdone is stored in the form of P.E.

∴ `P.E = 1/2 kx^2 = 1/2mω^2x^2`

Cases :

(i) at Mean position : - x = 0

∴ P.E = 0

(ii) at extreme position : x = ± a

∴ magnitude of P.E. is,

∴ `P.E =1/2 mω^2a^2`