Explanation Given information: Mass of block m = 3 l b Spring constant k = 2 l b / i n . Velocity v = 90 i n . / s For simple harmonic motion: x = x m sin ( ω n t + ϕ ) And, Natural frequency: ω n = k m Where k = 2 × 12 = 24 l b / f t ω n = 24 ( 3 32.2 ) ω n = 257.6 ω n = 16.04 r a d / s When the box is in equilibrium, x = 0 = x m sin ( 0 + ϕ ) ϕ = 0 x = ( 0.4673 ) sin ( 16.05 t ) x m = 7.5 16.05 x m = 0.4673 f t Now, x = ( 0.4673 ) sin ( 16.05 t ) _ _ _ _ _ _ ( 1 ) Then from the equation of simple harmonic motion: x = x m sin ( ω n t + ϕ ) o r , x = x m sin ( ω n t ) ∴ ϕ = 0 The amplitude at t = 0.90 s is: x = ( 0.4673 ) sin ( 16.05 × 0.90 ) x = ( 0

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 19.1, Problem 19.12P

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The position, velocity and acceleration of the block at t=0.90 s.

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The position, velocity and acceleration of the block at t = 0.90 s .

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