A particle is exceuting SHM of amplitude A .at what position from the mean position is the energy half potential and half kinetic?

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Solution

$A t t h i s p o i n t p o t e n t i a l e n e r g y = k i n e t i c e n e r g y \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} m ω^{2} (A^{2} - x^{2}) x^{2} = (A^{2} - x^{2}) 2 x^{2} = A^{2} x = \frac{\pm A}{\sqrt{2}} w h e r e x i s t h e d i s \tan c e f r o m m e a n p o s i t i o n .$

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A particle is exceuting SHM of amplitude A .at what position from the mean position is the energy half potential and half kinetic?

At this point potential energy = kinetic energy12mω2x2 = 12mω2A2-x2x2 =A2-x22x2 = A2x = ±A2where x is the distance from mean position.

$A t t h i s p o i n t p o t e n t i a l e n e r g y = k i n e t i c e n e r g y \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} m ω^{2} (A^{2} - x^{2}) x^{2} = (A^{2} - x^{2}) 2 x^{2} = A^{2} x = \frac{\pm A}{\sqrt{2}} w h e r e x i s t h e d i s \tan c e f r o m m e a n p o s i t i o n .$