Question

Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for the total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.

Answer 1

Consider a particle of mass m performing linear S.H.M. with amplitude A. The restoring force acting on the particle is F = −kx, where k is the force constant and x is the displacement of the particle from its mean position.

(1) Kinetic energy: At distance x from the mean position, the velocity is

v = ω `sqrt("A"^2-"x"^2)`

where ω = `sqrt("k"//"m")`. The kinetic energy (KE) of the particle is

KE = `1/2`mv² = `1/2` mω²(A² − x²)

= `1/2`k(A² − x²) ...........(1)

If the phase of the particle at an instant t is θ = ωt + x, where a is the initial phase, its velocity at that instant is

v = ωA cos (ωt + x)

and its KE at that instant is

KE = `1/2`mv² = `1/2`mω²A² cos²(ωt + x)   ....(2)

Therefore, the KE varies with time as cos²θ.

(2) Potential energy: Consider a particle of mass m, performing a linear S.H.M. along the path MN about the mean position O. At a given instant, let the particle be at P, at a distance x from O.

    Potential energy of a particle in SHM

The corresponding work done by the external agent will be dW = ( - F)dx =kx dx. This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.

∴ PE = `∫"dW"=∫_0^"x""kx dx"=1/2`kx² ............(3)

The displacement of the particle at an instant t being

x = A sin (ωt + x)

its PE at that instant is

PE = `1/2`kx² = `1/2`kA² sin²(ωt + x) ............(4)

Therefore, the PE varies with time as sin²θ

(3) Total energy: The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (2), total energy is
E = PE + KE

= `1/2`kx² + `1/2`k (A² − x²)

= `1/2`kx² + `1/2`kA² −`1/2`kx²

∴ E = `1/2`kA² = `1/2`mω²A² ......(5)

As m is constant, and w and A are constants of motion, the particle's total energy remains constant (or is conserved).