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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 0.2s. Find the acceleration and velocity of the body when the displacement is 5 cm.

Text Solution

Verified by Experts

Here `A=5cm, T=0.2s`
Velocity and acceleration at any displacement x are given by
`v=omegasqrt(A^(2)-x^(2))=(2pi)/(T)sqrt(A^(2)-x^(2))`
`a=-omega^(2)x=-(4pi^(2))/(T^(2))x`
When `x=5cm" "a=(2pi)/(0.2)sqrt(5^(2)-5^(2))=0`
`a=-(4pi^(2))/((0.2)^(2))xx5cms^(-2)=-500pi^(2)cms^(-2)=-5pi^(2)ms^(-2)`
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