At what distance from the mean position is the K.E. in simple harmonic

1.	At what distance from the mean position is the K.E. in simple harmonic oscillator equal to P.E.?
Answer» When the displacement of a particle executing S.H.M. is y, then its K.E. = \(\frac{1}{2}\)mω²(A² − y²) And P.E. = \(\frac{1}{2}m\omega^2y^2\) If K.E. = P.E. then, \(\frac{1}{2}m\omega^2(A^2\,-\,y^2)=\frac{1}{2}m\omega^2y^2\) Or 2y² = A² Or y = \(\frac{A}{\sqrt2}\)

At what distance from the mean position is the K.E. in simple harmonic oscillator equal to P.E.?

Answer»

When the displacement of a particle executing S.H.M. is y, then its

K.E. = \(\frac{1}{2}\)mω²(A² − y²)

And P.E. = \(\frac{1}{2}m\omega^2y^2\)

If K.E. = P.E.

then, \(\frac{1}{2}m\omega^2(A^2\,-\,y^2)=\frac{1}{2}m\omega^2y^2\)

Or 2y² = A²

Or y = \(\frac{A}{\sqrt2}\)

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At what distance from the mean position is the K.E. in simple harmonic oscillator equal to P.E.?

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