Question

Hi, I'm having trouble finding Theoretical Period T. For table 1, the mass is .068kg and pendulum length is 1.31m. For table 2, mass is 0.668kg and 25 degrees.

Answer 1

усе. SO Fs-mgen erogatos Expression for time period of simple Pendulery when the penduleury shifted from its equilibrium pon the restoring force ng sino causes penduleury to relturen back - Equithrine , to its & Origin pus? Hence the restoring force of masino displacement are in opposite in dir? Fs-mg sino Para produce e o art and for small o smo vo s-mge s-mg a -> a =-18h)* --SA for stm ( simple Hermonic Motion), acceleration 'a' obeys the relation with its displacement for small o. we can assume as; a=-wax tieplacement x2 nearly equel to are so compaining was 9/6 (ws 21/12) length Ins to +91 - 2016 I * Time period only depends Hence theoritically we can an inter length of string. salculatu cajculate time period using above formule. * Time period does not de peng on the wall of Table 1 & wheree g = 9.8 m/sec2 m 30.068 kg , L = 1.31 m 1.31 > 2 Y 3.14x/ = 2.296 sec. V 9.8 C) % erine Powritical time period - Exp. Avg. T 7 Time Perries) errey txot Ang T. x 100 213. 2.3-2.296 von 2. 296 2.296 2.296 2.35 2.296 xico 2.296 2.296 Period "Ts (Theositical Treenitical Theat Angle o 2.35 2.35 2.35 2.296 xico 2.296 2-36 2.296 2.36 - 2.29oxico 2296 Seanned with 2:37 CamScanner 2.296 2.37 2.296 2296 - Ico

Table: 2 ms o.fbsug ,0 525 ; Theontical T T2 lengur 1-31 Expt-Arg. I 2.36 2. 296. 12.36) 2. TUS 2.07 2.0556 0.94 1.98 1.9449 % coor Opt 2.36 - 2.296xico 2.296 2.07-2.0556 exico (2.0772 2.0556 1.98 -1.9449x160 2.9449 11-196 - 1.8165! xico (115) 1.8165 1.68-1.561 xico (1.602 1.56 0.82 1.188 1.8165 0.606 1.68 1.561 Cs Scanned with CamScanner