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Home » 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.
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.
CBSE | Class 11 | NCERT | Oscillations | Physics
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 mass of the particle executing simple harmonic motion is m. The particle’s displacement at a given time t is given by

x=A \sin \omega t

Velocity of the particle is given as \mathrm{v}=\mathrm{d} x / \mathrm{dt}=\mathrm{A} \omega \mathrm{cos} \omega \mathrm{t}

Instantaneous Kinetic Energy can be calculated as,

K=(1 / 2) m v^{2}

=(1 / 2) \mathrm{m}(\mathrm{A} \omega \cos \omega \mathrm{t})^{2}

=(1 / 2) m\left(A^{2} \omega^{2} \cos ^{2} \omega t\right)

Average value of kinetic energy over one complete cycle can be calculated as,

\begin{aligned} &K_{a v}=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m A^{2} \omega^{2} \cos ^{2} \omega t d t=\frac{m A^{2} \omega^{2}}{2 T} \int_{0}^{T} \cos ^{2} \omega t d t=\frac{m A^{2} \omega^{2}}{2 T} \int_{0}^{T} \frac{(1+\cos 2 \omega t)}{2} d t \\ &=\frac{m A^{2} \omega^{2}}{4 T}\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T} \\ &=\frac{m A^{2} \omega^{2}}{4 T}\left[(T-0)+\left(\frac{\sin 2 \omega t-\sin 0}{2 \omega}\right)\right] \\ &=\frac{1}{4} m A^{2} w^{2} \\ &\text { Average instantaneous potential energy, } U=(1 / 2) k x^{2}=(1 / 2) m \omega^{2} x^{2}=(1 / 2) m \omega^{2} A^{2} \sin ^{2} \omega t \end{aligned}

Average value of potential energy over one complete cycle can be calculated as

\begin{aligned} U_{a v} &=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega 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} \frac{(1-\cos 2 \omega t)}{2} d t \end{aligned}

=\frac{m \omega^{2} A^{2}}{4 T}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}

\begin{aligned} &=\frac{m \omega^{2} A^{2}}{4 T}\left[(T-0)-\frac{(\sin 2 \omega t-\sin 0)}{2 \omega}\right] \\ &=\frac{1}{4} m \omega^{2} A^{2} \end{aligned}

Kinetic energy = Potential energy

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