Explanation Given information: Mass of collar m = 5 Κ g Spring constant k = 720 Ν / m To simplify the analysis we measure the displacement from the position of static displacement of the spring, under the weight of the collar: Thus, static deflection is given by the equation: δ s t = W k δ s t = 5 × 9.81 720 δ s t = 0.068125 m Then, m x ¨ + k x = 0 5 x ¨ + 720 x = 0 Where, x is measured positively up from the position of static deflection. The equation of a simple harmonic motion is given as: x = x m sin ( ω n t + ϕ ) We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t), Since, v = d x d t v = d d t ( x m sin ( ω n t + ϕ ) ) v = ω n x m cos ( ω n t + ϕ ) Natural frequency ω n = k m ω n = 720 5 ω n = 12 r a d / s And, velocity of free fall of collar v o = 2 g h v o = 2 × 9.81 × 0.5 v o = 3.132 m / s Now, to calculate x m a n d ϕ , impose the conditions at impact and count the time from there. Thus, Free-fall time of collar: t 1 = 2 × 0.5 g t 1 = 1 9 To determine (b) The velocity of collar.

12th Edition

ISBN: 9781259977237

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 19.1, Problem 19.15P

To determine

(a)

The time period and motionof collar.

To determine

(b)

The velocity of collar.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19.1, Problem 19.14P

Chapter 19.1, Problem 19.16P

Students have asked these similar questions

A spring, with a 54 lb/in spring constant, is mounted on a vertical wall to support a 64.4 lb mass. At t=0 and an initial velocity of 9 in/sec. to the left, the mass is released from a position 2 inches to the right of the equilibrium position. Determine: a. The position x as a function of time. b. The maximum velocity c. The maximum acceleration

The recoil mechanisms of large artillery guns would benefit if designed well to allow their quickest return to the firing position without oscillation. A 1250 kg artillery gun is capable of firing 50 kg muzzles at a velocity of 250 km/h. The recoil mechanism is itself 52 kg in weight. The recoil system is to be designed in such a way that it can return to within 50 mm of its firing position 0.1 s after its maximum recoil. Determine (a) the stiffness of the recoil mechanism, (b) the damping coefficient of the recoil mechanism, and (c) the maximum recoil

Find the magnitude and direction of the velocity at point B.

Chapter 19 Solutions

Vector Mechanics For Engineers

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 - A 20-lb block is initially held so that the...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 - A 5-kg fragile glass vase is surrounded by packing...

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.

(a) The time period and motionof collar.

Solution Summary: The author explains that to simplify the analysis, we measure the displacement from the position of static displacement of the spring.