Explanation Calculation: Since ( c < c c ) which is under damped, the displacements x 1 , x 2 and x 3 may be assumed equal to maximum displacements. The expression for the differential equation of under damping as follows: x = x 0 e − ( c 2 m ) t sin ( ω 0 t + ϕ ) (1) Here, x is the displacement, x 0 is the initial displacement, c is the coefficient of viscous damping, m is the mass of the body, ω 0 is the frequency, t is the time, and ϕ is the phase angle. Apply boundary conditions at the maximum displacements. When t = t n then x = x n . Substitute t n for t , x n for x in the equation (1). x n = x 0 e − ( c 2 m ) t n sin ( ω 0 t n + ϕ ) Substitute 1 for sin ( ω 0 t n + ϕ ) x n = x 0 e − ( c 2 m ) t n ( 1 ) = x 0 e − ( c 2 m ) t n Apply boundary conditions at the next maximum displacements. When t = t n + k then x = x n + k . Substitute t n + k for t , x n + k for x in the equation (1). x n + k = x 0 e − ( c 2 m ) t n + k sin ( ω 0 t n + k + ϕ ) Substitute 1 for sin ( ω 0 t n + 1 + ϕ ) x n + k = x 0 e − ( c 2 m ) t n + k ( 1 ) = x 0 e − ( c 2 m ) t n + k The expression for the time at the maximum displacement and next maximum displacement as follows; ω D t n + k − ω D t n = k 2 π ω D ( t n + k − t n ) = k 2 π t n + k − t n = k 2 π ω D t n − t n + k = − k 2 π ω D Here, ω D is the damped circular frequency. Calculate the ratio of two successive displacements ( x n x n + k ) using the relation: Ratio of two successive displacement = x n x n + k Substitute x 0 e − ( c 2 m ) t n + k for x n + k and x 0 e − ( c 2 m ) t n for x n

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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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 19.130P

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Show that the logarithmic decrement can also expressed as (1k)ln(xnxn+k).

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Chapter 19.5, Problem 19.129P

Chapter 19.5, Problem 19.131P

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Chapter 19 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.2P Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - Prob. 19.6P Ch. 19.1 - Prob. 19.7P Ch. 19.1 - Prob. 19.8P Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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Show that the logarithmic decrement can also expressed as ( 1 k ) ln ( x n x n + k ) .

Solution Summary: The author explains that the logarithmic decrement can also be expressed as underset_(1k)mathrmln

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Show that the logarithmic decrement can also expressed as (1k)ln(xnxn+k).

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Chapter 19 Solutions