Ncert Exercises & Solutions

14.22 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.

$The equation of displacement of a particle executing SHM :$
$x = Asinωt$
$where,$
$A = Amplitude of oscillation$
$ω = Angular frequency = \sqrt{\frac{k}{M}}$
$The velocity of the particle is :$
$v = \frac{dx}{dt} = Aωcosωt$
$The kinetic energy of the particle is :$
$E_{k} = \frac{1}{2} {Mv}^{2} = \frac{1}{2} {MA}^{2} ω^{2} \cos^{2} ωt$
$The potential energy of the particle is :$
$E_{P} = \frac{1}{2} {kx}^{2} = \frac{1}{2} {Mω}^{2} A^{2} \sin^{2} ωt$
$For time period T, the average kinetic energy over a single$
$cycle :$
$\begin{array}{l} \begin{aligned} {(E_{k})}_{Avg} & = \frac{1}{T} \int_{0}^{T} E_{k} dt \\ = \frac{1}{T} \int_{0}^{T} \frac{1}{2} {MA}^{2} ω^{2} \cos^{2} ωtdt \\ = & \frac{1}{2 T} {MA}^{2} ω^{2} \int_{0}^{T} \frac{(1 + \cos 2 ωt)}{2} dt \end{aligned} \end{array}$
$\begin{aligned} = \frac{1}{4 T} {MA}^{2} ω^{2} {[t + \frac{\sin 2 ωt}{2 ω}]}_{0}^{T} \\ = \frac{1}{4 T} {MA}^{2} ω^{2} (T) \\ = \frac{1}{4} {MA}^{2} ω^{2} . . . (i) \end{aligned}$
$And, average potential energy over one cycle :$
$\begin{aligned} {(E_{P})}_{Avg} & = \frac{1}{T} \int_{0}^{T} E_{p} dt \\ = \frac{1}{T} \int_{0}^{T} \frac{1}{2} {Mω}^{2} A^{2} \sin^{2} ωtdt \\ = \frac{1}{2 T} {Mω}^{2} A^{2} \int_{0}^{T} \frac{(1 - \cos 2 ωt)}{2} dt \end{aligned}$
$\begin{array}{l} = \frac{1}{4 T} {Mω}^{2} A^{2} {[t - \frac{\sin 2 ωt}{2 ω}]}_{0}^{T} \\ = \frac{1}{4 T} {Mω}^{2} A^{2} (T) \\ = \frac{{Mω}^{2} A^{2}}{4} . . . (ii) \end{array}$

So the average kinetic energy for a given time period is equal to the average potential energy for the same time period.