Question

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A small bob is attached to a cord of length 1.2 m and is released from rest when A=5". Take d=0.75 m. 1.2 m С B A 4. Determine the time required for the bob to return to point A. (Round the final answer to three decimal places.) The time required for the bob to return to point Als

Answer 1

9.81ml Icon), 1.2 m 2 27 2.8592 ? ole -AB The pendulum moves betweens points A and B, the length of pendulum is L=LAB = 1.2m. g (en) 2 Time period LAB 27 3 = (con), (wm), = 2.8592 gadis = 2.1975 Sec The falling from A to B is one quarter Period givers by 4 ? TAB= (2.1975) CAB=0.54938 sec. The persduluns moves betweens points B ande, the length of the persdulum is l= LBC = 1.2m-d (=4BC = 1.2m-0.75m -0.45m ८ LBC (Wn2 9.81m/s 4.66904 0.45m 22 1.3457 sec (wm), = 4.66904 grad/s the motion from o to c and back to B is half period. ཟ་ g (cm) 27 (Wn) :72 27 - one 7888 = Ź az Ź (1.3457) CBCB = TBCB 0.67285 sec.

18 the persduluns moves B to A, lenath is 12m ZBA = 0.54938 sec the time required to return to A. + TBA 2 = 0.54938+0.67285 +0.54938 1.77161 Sec r no 7 = -AB + BCB 7 For path between A and B at point A maximum displacement (OA) is equal to om, . At point B the derivative of displacement and velocity is og = (cm)As ons VB = LABOR ; DA wn)AB OA VB las con heo og For path between B and C at point c maximum dieplacement (Oc) is equal to One. At point B OB = (Wn) Be Oma VB = LBC 6B VB = LBC (Wh) oc Oc G equate ③ LAB (WD) AB OA LBC (wn Bc Oc LAB (Wn), OA LBC (wn), oc 1.2 x 2.8592x5º = 0.45 * 4.66904 xoc Oc = 8.16999 OB = (WP) Be Oc