> Q. In a simple harmonic oscillator, at the mean position: – LIVE ANSWER TODAY

Q. In a simple harmonic oscillator, at the mean position:

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In a simple harmonic oscillator, at the mean position:

Solution:

Kinetic energy of particle of mass $m$ in SHM at any point is,

$=frac{1}{2} m omega^{2}left(a^{2}-x^{2}right)$

and potential energy $=frac{1}{2} m omega^{2} x^{2}$

where, $a$ is amplitude of particle and $x$ is the distance from mean position. So, at mean position, $x=0$

$KE =frac{1}{2} m omega^{2} a^{2}$ (maximum)

$PE =0$ (minimum)

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