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=0,30 m. 1.2 m @ce С B A 07 Determine the amplitude @ (Round the final answer to two decimal places.) The amplitude ec is

Answer 1

K OA = & 0.3m=d 1.2m OA 20 B c I 211 9 2. For the pendulum Given, Length of the Pendulum El-l2m d-0.3m Time period of a pendulum T= 2T 9 OC we know 1. For the pendulum between A&B The length of the pendulum LAB = 1.2m Now, Angular velocity 9 TAB 9.81 2.8592 rad's Wni = 2.8592 radls Llos Elt Eo 9 - W: 2 TI colo Et Wni Whi = 1.2 l-d The length of the pendulum = ABC ABC= 1.2.0.3 = 0.9m (BC=0.9m

Angular velocity W222 9 LBC 9.81 . 19 09 Wn2 = 10.9 whz - 3.3015 radls Now, consider the pendulum falling from A to B, then At A, Om: 0(0) (ie) Om= OA () At B, (we know) V=l So at B VB = LABOB As we know om = wom w - wni & Om = OA from from A to B → we have OB = om Wni OA (2) Velocity at B = VB = LABOB = 1BB (Wn, OA) Consider the pendulum rising from Bloc; (from Bloc l-lec) V= l A B , VB = LBC OB

do = VB loc from @ Vo LAB Con, OA LBC B = lA BWnio A oo LAB whi OA = ve Loc ömase IBC so we can write The maximum is found at C, angle Oc= Omas = Omane wna Substitute imane OC LAB wni OA LBC wna Oc = (1.2) (2.8592) 10.9) (3.3015) OC - 1.1547 OA [Given OA = 8 Oc = 1.1547 (50) Oc= 5.7735 The amplitude Oc= 5.77°