Program: National Engineering Technician Diploma Course code: ENSC 110D Class: Petroleum Lab Title: Pendulum with a yielding support Instructor: Mrs. Sharon Mohammed Full time Name: Kirn Johnson Student ID: 58605 Date: 28/10/2012 …show more content…
When y = 2.23s, x= 0 T2 = kd3+4π2lg
The equation is similar to y= mx + c Where y = 2.23s, x = 0, c= 4π2lg (l = 0.5m) m= 18.75m
Equation = T2 = kd3+4π2lg solving for g= 2.23 = 18.75 (0) + 4π2(0.5)g = 2.23 = 2π2g =2.23g = 19.74 g = 19.742.23 = 8.85ms-1
Gravity was calculated to be 8.85ms-1 The percentage error was calculated by using the formula: % error = ±.5mm value of d x 100. d = 0.24m 0.5mm to meters= 0.0005m
% error = ± 0.00050.24 x 100 = ± 0.0021 x 100 = 0.21%
The value of k, when compared to the equation of the straight line, is equal to the gradient , m. hence k is equal to18.75m.
Since the acceleration, g, was found to be 8.89ms-1 it can be seen that an error occurred whereas gravity to a free fall is said to be 9.81ms-1. As stated previously we can say that this error was caused due to the reaction time of the person recording the 20 oscillations. Question
Since T = 2secs and l = 1m By using the formula T2 = kd3+4π2lg where T2 = 4, k=18.75, l= 1m , g= 8.85ms-1 the gradient, m, is 18.75.
Substitute (2)2 = 18.75d3+ 4π2(1)8.85 * 4