Determining Height of Pendulum Attachment Point

[chikis]

Homework Statement

There was this question involving a pendulum expriment:

the five values of the square of the period (T^2) was plotted against the 5 values of the distance y in centimetre from the floor to the centre of the bob.

I was then asked to determine the intercept on the vertical axis and slope of the graph as well; I have determined the intercept and slope of the graph.

I was then asked to use the value of the intercept to determine the height H of the point of attachment of the pendulum to the floor. How do I go about that?

Homework Equations

The Attempt at a Solution

I have gotten the intercept and the slope of the graph as well.

 

[mishek]

You say that you know the distance from the floor Y and that you know the intercept (length of the rope)? Shouldn't the height of the point of attachment be H=y+L?

 

[chikis]

You say that you know the distance from the floor Y and that you know the intercept (length of the rope)? Shouldn't the height of the point of attachment be H=y+L?

Please what do you mean by 'L'?

 

[mishek]

L - the length of the rope.

 

[chikis]

L - the length of the rope.

I think I got the gist now:

H = l+y
->
l = H-y
the value of the intercept that I got from the actual graph that I plotted is 4.55 (sec)²

the period of oscillation of a simple pendulum is given as
T = 2(pi)/w
= 2(pi)(l/g)^1/2

since the intercept is at T²

squaring both sides

T²
=
(2(pi)(l/g)^1/2)²

we have
(91/20)²
=
4(pi)²(l/g)
=
4(pi)²(H-y/g)

assuming the acceleration due to gravity g is 10m/s and that the value of pi = 22/7

(91/20)²
=
4(22/7)²(H-y/10)
what do I subtitute in now as the value of the y?