(a) Explanation Given information: The mass of the motor ( M ) is 400 kg. The spring constant ( k ) is 150 kN / m . The damping constant ( c ) is 6500 N ⋅ s/m . The unbalanced mass of rotor ( m ) is 23 g. The distance from axis of rotation to center of the dashpot ( r ) is 100 mm. The speed of the rotor ( N ) is 800 rpm. Calculation: The system consists of four springs each with a spring constant of 150 kN / m Thus, the spring constant of the system is four times the spring constant of 150 kN / m . Calculate the spring constant of the system ( k ) using the relation: k = 4 k s p r i n g Here, k is the spring constant of the system is k and k s p r i n g is the spring constant of each spring. Substitute 150 kN / m for k s p r i n g . k = 4 ( 150 kN / m ) = 4 ( 150 × 10 3 N / m ) = 600 × 10 3 N / m Calculate the frequency of the periodic force ( ω f ) using the relation: ω f = 2 π N 60 Substitute 800 rpm for N . ω f = 2 π ( 800 rpm ) 60 = 83.776 rad / s The vibration of the system is due to the centrifugal force of the unbalanced mass of rotor. Calculate the magnitude of the periodic force ( P m ) using the formula: P m = m r ω f 2 Substitute 23 g for m , 100 mm for r and 83.776 rad / s for ω f . P m = ( 23 g ) ( 100 mm ) ( 83.776 rad / s ) 2 = ( 23 g × 1 kg 1000 g ) ( 100 mm × 1 m 1000 mm ) ( 83.776 ) 2 = ( 0.023 kg ) ( 0.1 m ) ( 83.776 ) 2 = 16.1424 N The expression for the differential equation for the forced damped vibration as follows: m x ¨ + c x ˙ + k x = P m sin ω f t (1) Here, x is the displacement of the system, x ˙ is the velocity, x ¨ is the acceleration and t is the time. The expression for the solution of the differential equation in equation (1) as follows: x = x m sin ( ω f t − ϕ ) (2) Here, x m is the maximum amplitude of the motion and ϕ is the phase difference. Calculate the velocity of the motion ( x ˙ ) using the relation: Differentiate equation (2) with respect to t . x ˙ = d x d t = d ( x m sin ( ω f t − ϕ ) ) d t = x m cos ( ω f t − ϕ ) × ω f = ω f x m cos ( ω f t − ϕ ) (3) Calculate the amplitude of forced damped vibration ( x m ) using the formula: x m = P m ( k − M ω f 2 ) 2 + ( c ω f ) 2 Substitute 16.1424 N for P m , 600 × 10 3 N / m for k , 400 kg for M , 83.776 rad / s for ω f , and 6 , 500 N ⋅ s / m for c (b) To determine The amplitude of the vertical motion ( x m ) of the motor.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

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, Problem 19.166RP

(a)

To determine

The amplitude of the fluctuating force (Fm) transmitted to the foundation.

(b)

To determine

The amplitude of the vertical motion (xm) of the motor.

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Chapter 19, Problem 19.165RP

Chapter 19, Problem 19.167RP

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Connect 1 Semester Access Card for 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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The amplitude of the fluctuating force ( F m ) transmitted to the foundation.

Solution Summary: The author calculates the spring constant of the system using the relation: k=4k_spring.

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The amplitude of the fluctuating force (Fm) transmitted to the foundation.

The amplitude of the vertical motion (xm) of the motor.

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Chapter 19 Solutions