Explanation Given information: The mass of the block is 2 kg , the spring constant is 128 N / m , the viscous damping coefficient is 0.6 N ⋅ s / m , the upward velocity of the block is 0.4 m / s . Expression for damping factor. ζ = c 2 k m ..... (I) Here, the damping factor is ζ , the viscous damping coefficient is c , the spring constant is k , the mass of the block is m . Expression for logarithmic decrement. δ = 2 π ζ 1 − ζ 2 ...... (II) Here, the logarithmic decrement is ζ . Calculation: Substitute 0.6 N ⋅ s / m for c , 2 kg for m , 128 N / m for k in Equation (I). ζ = 0.6 N ⋅ s / m 2 ( 2 kg ) ( 128 N / m ) = 0.6 kg ⋅ m ⋅ s / m ⋅ s 2 2 ( 2 kg ) ( 128 kg ⋅ m / m ⋅ s 2 ) = 0 To determine (b) The maximum upward displacement of the block from equilibrium after two cycles.

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ISBN: 9781259977237

Author: BEER

Publisher: MCG

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Question

Chapter 19.5, Problem 19.135P

To determine

(a)

The logarithmic decrement.

To determine

(b)

The maximum upward displacement of the block from equilibrium after two cycles.

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Chapter 19.5, Problem 19.134P

Chapter 19.5, Problem 19.136P

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(a) The logarithmic decrement.

Solution Summary: The author explains the logarithmic decrement, which is 0.118.