1.

Derive an expression for the total energy of a particle executing S.H.M. in terms of force constant (k).

Answer»

We know, potential energy of a particle executing S.H.M. (i.e., oscillator) at any instant is given by

P.E. = 1/2 Mω2x2   ...(i)

But ω = 2π/T = √{k/M}

or, Mω2 = k    ...(ii)

Hence eqn. (i) becomes

P.E. = 1/2 kx2    (iii)

Also, kinetic energy of the particle executing S.H.m is given by

K.E. = 1/2 Mω2(r2 - x2)

where r = amplitude of the particle

Using eqn. (ii), we get

K.E. = 1/2 k(r2 - x2)    ...(iv)

Thus the total energy at any instant of the particle executing S.H.M. is given by

E = K.E. + P.E. = 1/2 k(r2 - x2) + 1/2 kx2

or, E = 1/2 kr2    ....(v)


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