Figure TQ3.5 shows a simple pendulum of length li has a bob of mass mj. Anothersimple pendulum of length l, and having a bob of mass m; is suspended from m1.Show that the two natural frequencies of oscillation are given by the equationmị + m2 – 0-(.+ 1½\/m1 + m,\w²T2T2m2Figure TQ3.5Determine the natural frequencies of this system and compare the amplitude of m; andm, for each frequency. Work from first principles.1 = 1.0ml2 = 1.5mm; = 1kgm2 = 1.5kg Example C07.07butT,스 mag and T스mg+ T스 (Mi + ma)g(m, + m2)9,, + m(72 – 21)QuestionalsoA simple pendulum of length l; has a bob of mass mj. Another simple pendulum of(3)length lz and having a bob of mass m, is suspended from mį.Show that the two natural frequencies of oscillation are given by the equationand(4)14+1\/m, + m\o m1 + m2 -0Assuming that x, = a, cos ot and x, = a, cos ot, equation (3) becomes-m,a,o² cos ost = - m1 + m2)g(a, cos cot – a, cos ot)a, cos cot +m,, 41 – (m1 + m)g az _ m29(1mag/.i.e.(5)SolutionagLet the small angular displacements of the upper and lower pendula be 01 and 02and equation (4) becomesrespectively and let the corresponding linear displacements of mi and m, be xi and x2respectively, Figure Q7.7.-m,a,0? cos wt = -ma la, cos ot - a, cos wt)i.e.-1-4(6)Substituting for 4 in equation (5),az= M.i.e.-mla ++ m24+2\m, +1,1,i.e.+ m1 + m2 =0gFigure Q7.7NOTE : The ratio of the amplitudes may be obtained from equation (6)Then, if T, and Tz are the tensions in the two strings, the equation of motion of mị ism =-T,01+ T',0, .• (1)and the equation of motion of m, is- -T,0,(2)

Write the given expression. ω4g2-l1+l2l1l2m1+m2m1ω2g+m1+m2m1l1l2=0 Write the given data. m1=1…

Figure TQ3.5 shows a simple pendulum of length lh has a bob of mass m1. Another simple pendulum of length l; and having a bob of mass m; is suspended from mj. Show that the two natural frequencies of oscillation are given by the equation m, + m²\w² , m, + m2 _ 0 m2 Figure TQ3.5 Determine the natural frequencies of this system and compare the amplitude of m; and m, for each frequency. Work from first principles] m; = 1kg m; = 1.5kg 4 = 1.0m l, = 1.5m

Figure TQ3.5 shows a simple pendulum of length lh has a bob of mass m1. Another simple pendulum of length l; and having a bob of mass m; is suspended from mj. Show that the two natural frequencies of oscillation are given by the equation m, + m²\w² , m, + m2 _ 0 m2 Figure TQ3.5 Determine the natural frequencies of this system and compare the amplitude of m; and m, for each frequency. Work from first principles] m; = 1kg m; = 1.5kg 4 = 1.0m l, = 1.5m

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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