Q7. For the system shown in Figure 3 where a vertical mass m is connected to the midpoint of the slender bar of mass m and length L through a spring of stiffness k. Denote the displacement of the vertical mass by x and the small rotational angle of the slender bar by 0. Find the free vibration 0(t) x(t) response of the system with the following initial conditions: x(0) = 0, 0(0) = 0, x(0) = 2m/s,ė(0) = 0. Given: m = 2 kg, L = 1m, k = 1KN/m, the natural frequencies have been calculated to be w₁ = 16.69 rad/sec, w₂ = 36.69 rad/sec, the associate mode shapes are found to be 1 -0.2949 1 1.128 = M TX То slender bar
Q7. For the system shown in Figure 3 where a vertical mass m is connected to the midpoint of the slender bar of mass m and length L through a spring of stiffness k. Denote the displacement of the vertical mass by x and the small rotational angle of the slender bar by 0. Find the free vibration 0(t) x(t) response of the system with the following initial conditions: x(0) = 0, 0(0) = 0, x(0) = 2m/s,ė(0) = 0. Given: m = 2 kg, L = 1m, k = 1KN/m, the natural frequencies have been calculated to be w₁ = 16.69 rad/sec, w₂ = 36.69 rad/sec, the associate mode shapes are found to be 1 -0.2949 1 1.128 = M TX То slender bar
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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B7 Please help on the attached question.

Transcribed Image Text:Q7. For the system shown in Figure 3 where a vertical mass m is connected to the midpoint of the
slender bar of mass m and length L through a spring of stiffness k. Denote the displacement of the
vertical mass by x and the small rotational angle of the slender bar by 0. Find the free vibration
0(t)
x(t)
response
of the system with the following initial conditions: x(0) = 0, 0(0) = 0,
x(0) = 2m/s,ė(0) = 0. Given: m = 2 kg, L = 1m, k = 1KN/m, the natural frequencies have been
calculated to be w₁ = 16.69 rad/sec, w₂ = 36.69 rad/sec, the associate mode shapes are found to be
1
-0.2949
1
1.128
=
M
TX
То
slender bar
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