(a) Explanation Given Information: The mass of the fragile glass ( m ) is 5 kg. The box is dropped from the height ( h ) is 1 m. The acceleration due to gravity ( g ) value is 9.81 m/s 2 . Calculation: Calculate the velocity ( v ) at end of free fall using the relation: v = 2 g h Substitute 9.81 m/s 2 for g and 1 m for h . v = 2 ( 9.81 × 1 ) = 4.4294 m/s Show the given graph with slope as in Figure (1). Calculate the spring constant ( k ) by referring Figure 1: k = y x Substitute 100 N for y and 10 mm for x . k = 100 10 = 10 N/mm × 1000 mm 1 m = 10 , 000 N/m Calculate the natural circular frequency ( ω n ) using the relation: ω n = k m Substitute 10 , 000 N/m for k and 5 kg for m . ω n = 10000 5 = 44.721 rad/s Write the expression for the displacement ( x ): x = x m sin ( ω n t + ϕ ) (1) Calculate the angle ( ϕ ) : The time ( t ) and displacement ( x ) is zero when the velocity ( v ) is 4.4294 m/s (b) To determine The maximum acceleration ( a m ) the vase experience in g ’s.

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

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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,

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Chapter 19.1, Problem 19.10P

(a)

To determine

The amplitude (xm) of vibration for the vase.

(b)

To determine

The maximum acceleration (am) the vase experience in g’s.

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Chapter 19.1, Problem 19.9P

Chapter 19.1, Problem 19.11P

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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

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The amplitude ( x m ) of vibration for the vase.

Solution Summary: The author calculates the amplitude of vibration for the vase and the velocity at end of free fall using the relation.