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 + 1 then x = x n + 1 . Substitute t n + 1 for t , x n + 1 for x in the equation (1). x n + 1 = x 0 e − ( c 2 m ) t n + 1 sin ( ω 0 t n + 1 + ϕ ) Substitute 1 for sin ( ω 0 t n + 1 + ϕ ) x n + 1 = x 0 e − ( c 2 m ) t n + 1 ( 1 ) = x 0 e − ( c 2 m ) t n + 1 The expression for the time at the maximum displacement and next maximum displacement as follows; ω D t n + 1 − ω D t n = 2 π ω D ( t n + 1 − t n ) = 2 π t n + 1 − t n = 2 π ω D t n − t n + 1 = − 2 π ω D Here, ω D is the damped circular frequency. Calculate the ratio of two successive displacements ( x n x n + 1 ) using the relation: Ratio of two successive displacement = x n x n + 1 Substitute x 0 e − ( c 2 m ) t n + 1 for x n + 1 and x 0 e − ( c 2 m ) t n for x n

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Author: BEER

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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 19.129P

To determine

Show that the ratio of any two successive maximum displacement (xn) and (xn+1) is a constant and that the natural logarithm of this ratio called the logarithmic decrement is

lnxnxn+1=2π(c/cc)1−(c/cc)2.

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Chapter 19.5, Problem 19.128P

Chapter 19.5, Problem 19.130P

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Show that the ratio of any two successive maximum displacement ( x n ) and ( x n + 1 ) is a constant and that the natural logarithm of this ratio called the logarithmic decrement is ln x n x n + 1 = 2 π ( c / c c ) 1 − ( c / c c ) 2 .

Solution Summary: The author demonstrates that the ratio of any two successive maximum displacement (x_n) is a constant, and the natural logarithm of this ratio called the 'loga

Concept explainers

Show that the ratio of any two successive maximum displacement (xn) and (xn+1) is a constant and that the natural logarithm of this ratio called the logarithmic decrement is

lnxnxn+1=2π(c/cc)1−(c/cc)2.

Want to see the full answer?

Chapter 19 Solutions

Show that the ratio of any two successive maximum displacement ( x n ) and ( x n + 1 ) is a constant and that the natural logarithm of this ratio called the logarithmic decrement is ln x n x n + 1 = 2 π ( c / c c ) 1 − ( c / c c ) 2 .

Solution Summary: The author demonstrates that the ratio of any two successive maximum displacement (x_n) is a constant, and the natural logarithm of this ratio called the 'loga

Concept explainers

Show that the ratio of any two successive maximum displacement (xn) and (xn+1) is a constant and that the natural logarithm of this ratio called the logarithmic decrement is lnxnxn+1=2π(c/cc)1−(c/cc)2.

Want to see the full answer?

Chapter 19 Solutions

Show that the ratio of any two successive maximum displacement (xn) and (xn+1) is a constant and that the natural logarithm of this ratio called the logarithmic decrement is

lnxnxn+1=2π(c/cc)1−(c/cc)2.