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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.
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Solution
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).
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