Find Error in Slope Using Least Squares Method: Simple Pendulum Experiment”

APPARATUS REQUIRED

i) Simple pendulum

ii) Stop watch

iii) Meter scale

iv) Clamp and thread

THEORY

The principle of Least Squares:

least-square is one of the most commonly used methods in numerical computation. Essentially it is a technique for solving a set of equations where there are more equations than unknowns ie. an overdetermined set of equations. This set of notes shows the origins of a particular form of the algorithm, batch linear least-squares.

The most important application is in data fitting. The best fit in the least squares sense minimizes the sum of squared residuals, a residual being the difference between an observation value and the fitted value provided by a model. When the problem has substantial uncertainties

In the independent variable, Then simple regression and least square methods leave problems: in such cases, the methodology required for fitting errors in variable models may be considered instead of that for least squares. 

Least squares depending problem fall into two categories,

Linear or ordinary least squares and non-linear least squares depending on whether or not the residuals are linear are all unknown. This problem occurs in statistical regression analysis; it has a closed-form solution. A closed form solution is any formula that can be evaluated in a finite number of standard operations. The non-linear problem has no closed form solution and is  usually solved by iterative refinement at which iteration is approximated by a linear one, and thus the core calculation is similar in both cases.

A residual is defined as the difference between the actual value of the dependent variable and the value predicted by the model.

γ_i = y_i – f(x_i, β)

An example of a model is that of the straight line in two dimensions. Denoting the intercept as β_o and the slope as β₁ the model function is given by

                          f(x, β) = β_o + β₁x

let, the equation of straight line y = a+bx – – – – – – (i)

which is fitted to given points (x₁y₁) (x₂ y₂) (x₃ y₃)- – – – – – – – (x_n y_n)

let R, be the theoretical value for x₁ than, E₁ = y₁-R₁ = y₁– (d+ bx₁)

On solving equation (iii) we get the value of ‘a’ and ‘b’ and putting this value in equation (i) we get the value of equation in line of best fit.

OBSERVATION

Diameter of ball(d) = M.SR + C.SR x L.C.

       = 18+50× 0.01  =18.50m

r=d/2  = 18.50/2  = 9.25cm  = 0.925m

Time for 20 Oscillation	Time forPer oscillation	Effectivelength(l)	g(ms2)4𝝅2l/T2	Mean(g)	T2
17	17/20 =0.85	20	195.11		0.72
25	25/20 =1.25	40	1011.24		1.56
31	31/20 = 1.55	60	985.96	1021.95	2.40
35	35/20 = 1.75	80	1031.07		3.06
40	40/20 = 2	100	985.96		4

Slope of (m) =g/4𝝅²  =  1021.95/ 4×(3.14)² = 25.92 cms^-2

Time for 20 Oscillation	Time for 1 Oscillation	T2	Xi X1+X2+- –	x̄	Effective length(l)	Yi Y1+y2+- –	Ȳ
17	0.05	0.75			20
25	1.25	1.56	11.74	2.348	40
31	1.55	2.40			60	300	60
35	1.75	3.06			80
40	2	4			100

which is compared with equation of straight line y= mx

If l= 20 then l= 25.92 × 0.72 = 18.7

   l= 40 then l= 25.92 × 1.56 = 40.4

   l= 60 then l= 25.92 × 2.4 = 62.2

   l= 80 then l= 25.92 × 3.06 = 79.31

   l= 100 then l= 25.92 × 4 = 103.68

RESULT:

Thus the slope m= 25.92 and acceleration due to gravity (g) = 10.21 m/sec²

PERCENTAGE ERROR

Observed value of (o)g= 10.21 m/s²

standard value of (s)g = 9.8 m/s²

percentage errors=| 9.8-10.21/9.8 ×100%|

    = 4.32

CONCLUSION:

Hence, the acceleration due to gravity can be determined by using a simple pendulum.