Consider a single-degree-of-freedom system with the following equation of motion. mä+ci+kx = 0 In free vibration, the response of the system is measured experimentally, as shown in the figure below. The equivalent spring constant is k = 5 N/m. Successive peak displacements are ₁ = 2 mm and 2 = 1 mm. If the period of oscillation of the system is Ta = 1 s, determine the mass of the system. ooooo 112.6158 kg 15.6411 kg 0.1251 kg 3.3785 kg 56.0576 kg X1 A Ta X2

Consider a single-degree-of-freedom system with the following equation of motion. mä+ci+kx = 0 In free vibration, the response of the system is measured experimentally, as shown in the figure below. The equivalent spring constant is k = 5 N/m. Successive peak displacements are ₁ = 2 mm and 2 = 1 mm. If the period of oscillation of the system is Ta = 1 s, determine the mass of the system. ooooo 112.6158 kg 15.6411 kg 0.1251 kg 3.3785 kg 56.0576 kg X1 A Ta X2

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

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. Generally, there are three types of vibration: longitudinal vibration, transverse vibration, and torsional vibration.

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Consider a single-degree-of-freedom system with the following equation of motion.
In free vibration, the response of the system is measured experimentally, as shown in the figure below. The equivalent spring constant is k = 5 N/m. Successive peak
displacements are *₁ 2 mm and 2 = 1 mm. If the period of oscillation of the system is Ta = 1 s, determine the mass of the system.
=
0 0 0 0 0
112.6158 kg
15.6411 kg
0.1251 kg
3.3785 kg
56.0576 kg
mx + cx + kx = 0
X
X1
x2
A
Td
t

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