Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
The equation of displacement of a particle executing SHM at an instant t is given as:
x = Asin ωt
where,
A = Amplitude of oscillation
ω = Angular frequency = √k/M
The velocity of the particle, v = dx/dt = Aωcosωt
The kinetic energy of the particle is,
Ek = 1/2 Mv2 = 1/2 MA2 ω2 cos2 ωt
The portential energy of the particle is,
Ep = 1/2 kx2 = 1/2 M2 ω2 A2 sin2 ωt
For time period T, the average kinetic energy over a single cycle is given as:
Average potential energy over one cycle is given as,
From equations (i) and (ii) we can say that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.
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