Question

A small bob is attached to a cord of length 1.2 m and is released from rest when A= 5º. Take d = 0.60 m. 1.2 m 001 с B A

Answer 1

Solution d=06m -> DAR while for the path -1.2m cordis but the time period required is only for the path A to B and then from Civen, dength of the Cord l= 12m ,on 5 For the Osci lections of pendulum , the length of the cord is not the same for the entire Cycle. For the path between points A & B the length of cand is 1.2m between B&c the dength of dge = (1+2-d)m > Natural frequency for path between A &B g 9.81 MAR (cm) Time period T= (On) AB B to A. 20 & Helf the time period of between point a fB. (Won AB d. 8592 rad [sec This is the time period taken to complete one full oscillation, is the time period for the path con las 2T 279 z iconimo 2.8592 TAB 1.09876 sec. loc = (1.2-d) = (1-2 -0.6) 1 abc -0.6 m = 4.0435 -rad/sec - (On) BC (walbe 0.6 be

Time period T= 1 21 [ Corresponding to koniec (von Be It is the time for one full oscillation, but the time required is only for path B toc and then from cto B. i. Half the time pertod of is the time period for the path (Wh) Be between points B &c. loc: $ con ] 1/4 au 20 41043) = 0.77695 sec Toc The time period to return to A t = TAB & TBC - 1.09876)+(0.77595) ta = 1.876 sec : The time Olemised for the bob to return to paint Ais t = 1.876 secling Now For path between A & B.- At Air - The marimum displacement in the path between A dB as at A is the marimer displacement i en The displacement OA = 0m 1 ALB The derivatne of diphrement at B ê, kan het og - (won) A velocity at B

(1.2) (2.8592) (59) - (0.6) (4.0435) Oc The time to required to the bob to return to point. A is For the path between point B & C : at cl Maximum displacement between B &c as One- The displacement at C is maximum i. e, Oma at B : the desirative of displacement at B (wn). Oma Og = (wonloc Oc Velocity at B "B. = loc (Wonde de " Requete 2 & ③ - dAS (LONA PA = loe (conlec Oc OC 707° T 1- 876 See The amplitude foc 7.670 please like the Answer, Thank you