Explanation Given information: The weight of the machine ( W ) is 800 lb. The spring constant ( k ) is 200 lb/inch. The coefficient of damping ( c ) is 8 lb ⋅ s / in . . The periodic force is applied with a frequency ( f f ) is 2.5 cycles. The magnitude of the periodic force ( P m ) is 30 lb. Calculation: Convert the damping coefficient ( c ) from lb ⋅ s / in . to lb ⋅ s / ft : c = ( c ) lb ⋅ s / in . × lb ⋅ s 1 in . × 1 ft 12 in . Here, ( c ) lb ⋅ s / in . is the damping coefficient in lb ⋅ s / in . . Substitute 8 lb ⋅ s / in . for c . c = 8 × lb ⋅ s 1 in . × 1 ft 12 in . = 96 lb ⋅ s / ft Calculate the mass of the machine ( m ) element using the formula: m = W g Here, g is the acceleration due to gravity. Substitute 800 lb for W and 32.2 ft / s 2 for g . m = 800 lb 32.2 ft / s 2 = 24.845 lb ⋅ s 2 / ft The system consists of two springs each with a spring constant of 200 lb / in . Thus, the spring constant of the system is given by two times the spring constant of 200 lb / in . . Calculate the spring constant of the system ( k ) for two springs using the relation: k = 2 k s p r i n g Here, k is the spring constant of the system and k s p r i n g is the spring constant of each spring. Substitute 200 lb / in . for k s p r i n g . k = 2 ( 200 lb / in . ) = 400 lb / in . = 400 lb 1 in . × 1 ft 12 in . = 4800 lb / ft Calculate the natural circular frequency ( ω n ) using the formula: ω n = k m Substitute 4800 lb / ft for k and 24.845 lb ⋅ s 2 / ft for m . ω n = 4800 lb / ft 24.845 lb ⋅ s 2 / ft = 13

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Forced undamped vibrations

In mechanical engineering, vibration is expressed as the term that occurs due to an object's backward and forward movement about a point of equilibrium. Generally, there are three types of vibrations which are,

Videos

Question

Chapter 19.5, Problem 19.139P

To determine

The amplitude (xm) of the steady state vibration of the element.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19.5, Problem 19.138P

Chapter 19.5, Problem 19.140P

Students have asked these similar questions

I do not understand how to approach this question. I tried to answer it but I kept getting it incorrect.

(read image)

Chapter 19 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The amplitude ( x m ) of the steady state vibration of the element.

Concept explainers

Videos

The amplitude (xm) of the steady state vibration of the element.

Want to see the full answer?

Chapter 19 Solutions