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Mechanical Properties Of Fluids

Question
CBSEENPH11019165

Show that for a particle in linear simple harmonic motion, the average kinetic energy over a period of oscillation is half the total energy.

Solution
Let a particle of mass m execute simple harmonic motion with frequency ω. Let a be the amplitude of vibration.
space therefore space space space Total space energy space of space particle comma space straight E equals 1 half mω squared straight A squared
The position of particle at any instant during simple harmonic motion is given by,
space straight x space equals space straight A space s i n space omega t
therefore space thin space K i n e t i c space e n e r g y space straight K space equals 1 half m v squared

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half m omega squared straight A squared c o s squared omega t

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight E space c o s squared omega t 

Now the average kinetic energy of particle over complete cycle is,
top enclose straight K space equals 1 over straight T integral presubscript 0 presuperscript straight T straight E space cos squared space ωt space dt space

space space space space equals straight E over straight T integral presubscript 0 presuperscript straight T fraction numerator straight I plus cos 2 ωt over denominator 2 end fraction dt space

space space space space equals fraction numerator straight E over denominator 2 straight T end fraction open square brackets straight t plus fraction numerator sin space 2 space ωt over denominator 2 straight omega end fraction close square brackets subscript 0 superscript straight T space

space space space space equals fraction numerator straight E over denominator 2 straight t end fraction left square bracket straight T right square bracket equals straight E over 2
Hence proved.