Explanation Given information: The weight of block A ( m ) is 18 lb. The spring constant ( k ) is 10 lb/in. The magnitude of periodic force ( δ ) is δ m sin ω f t . The deflection of the spring ( δ m ) is 6 in. The force exerted by the spring on the block is less than 30 lb. Calculation: Convert the unit of spring constant from lb / in . to lb / ft : k = 10 lb / in . = 10 lb 1 in . × 1 ft 12 in . = 120 lb / ft Calculate the mass of the block A ( m ) using the formula: m = W g Here, g is the acceleration due to gravity. Substitute 18 lb for W and 32.2 ft / s 2 for g . m = 18 lb 32.2 ft / s 2 = 0.559 lb ⋅ s 2 / ft Calculate the natural circular frequency of vibration ( ω n ) using the formula: ω n = k m Substitute 120 lb / ft for k , and 0.559 lb ⋅ s 2 / ft for m . ω n = 120 lb / ft 0.559 lb ⋅ s 2 / ft = 14.652 rad / s Calculate the amplitude of forced vibration ( x m ) using the formula: x m = δ m 1 − ( ω f 2 ω n 2 ) 2 (1) Here, δ m is the deflection of the spring and ω f is the frequency of the periodic force. Calculate the periodic force ( F m ) using the relation: F m = − k ( x m − δ m ) (2) Substitute equation (1) in equation (2). F m = k ( x m − δ m ) = k ( ( δ m 1 − ( ω f / ω n ) 2 ) − δ m ) = k δ m ( 1 1 − ( ω f / ω n ) 2 − 1 ) = k δ m ( ( ω f / ω n ) 2 1 − ( ω f / ω n ) 2 ) Substitute 120 lb / ft for k , and 6 in. for δ m . F m = ( 120 lb / ft ) ( 6 in . × 1 ft 12 in . ) ( ( ω f / ω n ) 2 1 − ( ω f / ω n ) 2 ) = ( 120 ) ( 0.5 ft ) ( ( ω f / ω n ) 2 1 − ( ω f / ω n ) 2 ) = 60 ( ( ω f / ω n ) 2 1 − ( ω f / ω n ) 2 ) (3) For in-phase motion: The spring force is less than 30 lb

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 undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Gen…

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Chapter 19.4, Problem 19.105P

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Chapter 19.4, Problem 19.104P

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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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