Explanation Given information: Mass of bar AB, m = 5 Κ g Mass of collar m C = 2.5 Κ g Length of bar AB, l = 0.6 m Spring constant, k = 1.5 Κ Ν / m Concept used: By using the principle of conservation of energy for the free vibration of un-damped system the energy is partly kinetic and partly potential, keeping the total energy being constant. Thus, T + V = Constant Kinetic energy of the system T = 1 2 [ m l 2 3 + m C l 2 ] θ ˙ 2 And, Potential energy of the system V = k ( l θ ) 2 2 + m g l θ 2 4 + m C g l θ 2 2 Calculation: Total energy = kinetic energy + Potential energy 1 2 [ m l 2 3 + m C l 2 ] θ ˙ 2 + k ( l θ ) 2 2 + m g l θ 2 4 + m C g l θ 2 2 = 0 Now, differentiating the above equation with respect to time (t) to find the natural frequency. Thus, d d t [ 1 2 [ m l 2 3 + m C l 2 ] θ ˙ 2 + k ( l θ ) 2 2 + m g l θ 2 4 + m C g l θ 2 2 ] = 0 d d t [ 1 2 [ m l 2 3 + m C l 2 ] θ ˙ 2 ] + d d t [ 2 k l 2 θ 2 + m g l θ 2 + 2 m C g l θ 2 4 ] = 0 1 2 [ m l 2 3 + m C l 2 ] d d t [ θ ˙ 2 ] + [ 2 k l 2 + m g l + 2 m C g l 4 ] d d t [ θ 2 ] = 0 1 2 [ m l 2 3 + m C l 2 ] 2 θ ¨

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

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 19.3, Problem 19.82P

To determine

The period of oscillation of the resulting vibrating motion.

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Chapter 19.3, Problem 19.81P

Chapter 19.3, Problem 19.83P

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The period of oscillation of the resulting vibrating motion.

Solution Summary: The author explains the principle of conservation of energy for the free vibration of un-damped system.

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The period of oscillation of the resulting vibrating motion.

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