Show that for a particle performing linear simple harmonic motion…

Question

Show that for a particle performing linear simple harmonic motion, the average kinetic energy of any period of oscillation is equal to the average potential energy of the same period.

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Answer

Suppose a particle of mass performs simple harmonic motion for time period T. At the instant when time is measured from the mean position, the displacement of the particle is
$y=a \sin \omega t\ldots\ldots (1)$
We know that
$\begin{aligned}v & =\text { Velocity of particle } \\& =\frac{d y}{d t}=\frac{d}{d t}(a \sin \omega t) \\v & =a \omega \cos \omega t \ldots\ldots(2) \\KE, E_k & =\frac{1}{2} m v^2=\frac{1}{2} m(a \omega \cos \omega t)^2 \\& =\frac{1}{2} m a^2 \omega^2 \cos ^2 \omega t \\\text { P.E., } E_p & =\frac{1}{2} ky^2=\frac{1}{2} k(a \sin \omega t)^2 \\E_p & =\frac{1}{2} ka^2 \sin ^2 \omega t \\& =\frac{1}{2} m \omega^2 a^2 \sin ^2 \omega t \\\left(E_k\right)_{a v} & =\text {Average kinetic energy of one cycle} \\& =\frac{1}{T} \int_0^{T} E_k d t \\& =\frac{1}{T} \int_0^{T}\left(\frac{1}{2} m a^2 \omega^2 \cos { }^2 \omega t\right) d t\end{aligned}$
$\begin{aligned}& =\frac{1}{2 T} m a^2 \omega^2 \int_0^{T} \cos ^2 \omega t d t \\& =\frac{1}{2 T} m a^2 \omega^2 \int_0^{T}\left(\frac{1+\cos 2 \omega t}{2}\right) d t \\& =\frac{1}{4 T} m a^2 \omega^2\left[\int_0^{T} 1 \cdot d t+\int_0^{T} \cos 2 \omega t d t\right] \\& =\frac{1}{4 T} m a^2 \omega^2\left[(T-0)+\left(\frac{\sin 2 \omega t}{2 \omega}\right)_0^{T}\right] \\& =\frac{ma^2 \omega^2}{4 T}\left[T+\frac{1}{2 \omega}\left(\sin \frac{4 \pi}{T} \times T-\sin 0\right)\right] \\& =\frac{ma^2 \omega^2}{4 T}\left[T+\frac{1}{2 \omega}(0-0)\right] \\\therefore \sin n \pi & =0, n=0,1,2 \ldots \\\left(E_i\right)_{a v} & =\frac{1}{4} ma^2 \omega^2\ldots\ldots (3)\end{aligned}$
Average potential energy one cycle
$\begin{array}{l}\left(E_{p}\right)_{a v}=\frac{1}{T} \int_0^{T} E_{p} \cdot d t \text { given by } \\\left(E_{p}\right)_{av}=\frac{1}{T} \int_0^{T} \frac{1}{2} m \omega^2 a^2 \sin ^2 \omega t d t \\=\frac{m \omega^2 a^2}{2 T} \int_0^{T} \sin ^2 \omega t d t \\=\frac{m \omega^2 a^2}{2 T} \int_0^T\left(\frac{1-\cos 2 \omega t}{2}\right) d t \\\because \cos 2 \theta=1-2 \sin ^2 \theta \\=\frac{1}{4 T} m \omega^2 a^2\left[\int_0^{T} 1 . d t-\int_0^{T} \cos 2 \cos d t\right] \\=\frac{1}{4 T} m \omega^2 a^2\left[(T-0)-\left(\frac{\sin 2 \omega t}{2 \omega}\right)_0^{T}\right] \\\left(E_{p}\right)_{a v}=\frac{1}{4} m a^2 \omega^2\ldots\ldots (4)\end{array}$
Thus, from equation (3) and (4) we see that the average K.E. over one oscillation time period is and during the same period the average P.E. is equiralent to.