Explanation Given that, X m = 0.05 in Given that, Weigh, w = 800 lb ⥤ m = 500 32.2 = 15.527 lb.s 2 /ft P m = 20 lb ƒ = 2 Hz Damping coefficient, c = 10 lb.s/in = 120 lb.s/ft k = 2 × 250 = 500 lb/in = 6000 lb/ft Undamped natural frequency is given by: ω n = k m = 6000 15.527 = 19.65 rad/s and f n = ω n 2 π = 19.65 2 = 3.127 Hz Now the critical damping coefficient is given by: C c = 2 m ω n = 2 × 15.257 × 19.65 = 610.21 lb.s/ft Amplitude of the steady-state vibration of the element is given by: X m = P m k ( 1 − ( ω ω n ) 2 ) 2 + [ 2 ( c c c ) ( ω ω n ) ] 2 ---------------- (Eq

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Free Damped Vibrations

Free vibration is a type of vibration in which the external force is removed after giving the initial displacement to a body/system. In other words, in this type of vibration, force is applied at the initial displacement of the system, and after the init…

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Chapter 19.5, Problem 19.140P

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The required value of the coefficient of damping if the amplitude of the steady-state vibration of the element is to be 0.05 in.

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Chapter 19.5, Problem 19.139P

Chapter 19.5, Problem 19.141P

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The required value of the coefficient of damping if the amplitude of the steady-state vibration of the element is to be 0.05 in.

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The required value of the coefficient of damping if the amplitude of the steady-state vibration of the element is to be 0.05 in.

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