(a) Explanation Given information: The force deflection equation for a nonlinear spring fixed at one end is F = 2.5 x 1 / 2 . The weight of the block ( W ) is 6 oz. Assume the acceleration due to gravity ( g ) is 32.2 ft/s 2 . Calculation: The equation for nonlinear spring fixed at one: F = 2.5 x 1 / 2 (1) Convert ounce into lb. W = 6 oz × 1 lb 16 oz = 0.375 lb Here, the force is cause by the weight of the block ( W ). F = W Calculate the deflection ( x 0 ) : Rewrite the equation (1), W = 2 (b) To determine The frequency ( f n ) of vibration of the block.

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977251

Author: BEER

Publisher: MCG

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Chapter 19.1, Problem 19.32P

(a)

To determine

The static deflection (x0) if a 6 oz block is suspended from the spring.

(b)

To determine

The frequency (fn) of vibration of the block.

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Chapter 19.1, Problem 19.31P

Chapter 19.1, Problem 19.33P

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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 - A particle moves in simple harmonic motion....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 - A simple pendulum consisting of a bob attached to...Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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The static deflection ( x 0 ) if a 6 oz block is suspended from the spring.

Solution Summary: The author explains the force deflection equation for a nonlinear spring fixed at one end.

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The static deflection (x0) if a 6 oz block is suspended from the spring.

The frequency (fn) of vibration of the block.

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Chapter 19 Solutions