Question

Derive an expression for the total energy of a particle executing S.H.M.

Answer 1

A particle executing simple harmonic motion possesses both the potential energy and kinetic energy. The potential energy is due to the displacement of the particle from mean position and the kinetic energy is due to the motion of the particle. At the mean position, the displacement is zero and hence the potential energy of the particle at the mean position is also zero. On the other hand, at mean position, velocity is maximum and hence the kinetic energy of particle is maximum at this position. At extreme position, potential energy is maximum because of maximum displacement and kinetic energy is zero due to zero velocity of the particle. At the intermediate positions, the energy of the particle is the sum of kinetic energy and potential energy.

i.e., Energy, E = P.E. + K.E.    ...(i)

Potential Energy:- Consider a particle of mass M executing S.H.M. Let the particle is displaced through a distance x from mean position O to position P (See fig).

The acceleration of the particle at P is a = -ω²x and is directed towards mean position. Therefore, the restoring force acting on the particle is

F = Ma = Mω²x

Let the particle is displaced through a distance dx to the Q, therefore, work done on the particle is

dW = vector(F.dx)

= F dx cos 180°   (∴ θ = 180°)

= -F dx

Putting the value of F, we get

dW = M ω²x dx    ...(ii)

Total work done to displace the particle from mean position to point P, where displacement = x can be calculated by integrating eq. (ii)

This work done is stored as the potential energy of the particle.

Thus the potential energy of the particle is

P.E. = W = 1/2 Mω²x²    ....(iii)

Kinetic energy:- The kinetic energy of the particle executing S.H.M. at any instant is given by

K.E. = 1/2 Mv²

But, velocity of a particle executing S.H.M. is given v = ω√{r² - x²}

where r is the amplitude of the particle.

Hence, K.E. = 1/2 Mω² (r² - x²)   ....(iv)

Substituting the values of eqn. (iii) and (iv) in eqn. (i) we get

E = 1/2 Mω²x² + 1/2 Mω²(r² - x²)

or, E = 1/2 Mω²r²   ...(v)

Since ω = 2πv

We have, E = 1/2 M x 4π²v²r²

or, E = 2π² Mv²r²    ....(vi)

Thus, energy of the particle executing S.H.M. is directly proportional to

(i) square of the amplitude (r) of the particle and (ii) square of the frequency (v) of the vibrating particle.