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A body describes simple harmonic motion ...

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is 3 cm.

Text Solution

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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)`
Whem `y=3" cm"=0.03"m "rArra=(10pi)^(2)xx0.03=3pi^(2)" m s"^(-2)`.
`v=10pixxsqrt((0.05)^(2)-(0.03)^(2))rArrv=10pixx0.04=0.4pi" ms"^(-1)`.
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