find the ratio of kinetic energy and potential energy of a particle executing simple harmonic motion at midpoint between the mean position and extreme position
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In a SHM of an object, based on the principle of Conservation of Energy we have at any instant : KE + PE = constant = (1/2) K A²
[(1/2) K A² ] is the maximum PE when the motion reaches its maximum amplitude A , At which instant its KE = 0 ,K is elastic or restoring Force constant.
We will take PEx = (1/2) K x² and KE = (1/2) m v²
∴ (1/2) m v² + (1/2) K x² = (1/2) K A²
When x = A/2 we have
(1/2) m v² + (1/2) K (A/2)² = (1/2) K A² ⇒ KE + (1/8)KA² = (1/2) K A²
∴KE = (3/8)KA² , Therefore the ratio of KE/PE at this instant
= [(3/8)KA²] / [(1/8)KA²] = 3:1
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In a SHM of an object, based on the principle of Conservation of Energy we have at any instant : KE + PE = constant = (1/2) K A²
[(1/2) K A² ] is the maximum PE when the motion reaches its maximum amplitude A , At which instant its KE = 0 ,K is elastic or restoring Force constant.
We will take PEx = (1/2) K x² and KE = (1/2) m v²
∴ (1/2) m v² + (1/2) K x² = (1/2) K A²
When x = A/2 we have
(1/2) m v² + (1/2) K (A/2)² = (1/2) K A² ⇒ KE + (1/8)KA² = (1/2) K A²
∴KE = (3/8)KA² , Therefore the ratio of KE/PE at this instant
= [(3/8)KA²] / [(1/8)KA²] = 3:1
HOPE IT HELPS YOU ❤️❤️
PLEASE MARK THIS AS THE BRAINLIEST ❤️❤️
pranamsinha:
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