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A body describes simple harmonic with an...

A body describes simple harmonic with an amplitude of 5 cm and a period of `0.2` s Find the acceleration and velocity of body when the displacement is 5 cm

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We are given : `A=5" cm"=0.05"m",T=2s`
We are to calculate : acceleration and velocity of body.
Using relation : `a=omega^(2)y" and "v=omegasqrt(A^(2)-y^(2))rArromega=(2pi)/(T)=(2pi)/(0.2)=10pi" rad s"^(-1)`
When `y=5` cm `=0.05"m "rArra=omega^(2)y=(10pi)^(2)xx0.05=5pi^(2)" m s"^(-2)`.
`v=omegasqrt(A^(2)-y^(2))=10pisqrt((0.05)^(2)-(0.05)^(2)):.v=0`.
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