# Pendulum bulb expriment

1. Oct 23, 2012

### chikis

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. The attempt at a solution
I have gotten the intercept and the slope of the graph as well.

2. Oct 23, 2012

### 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?

3. Oct 24, 2012

### chikis

Please what do you mean by 'L'?

4. Oct 24, 2012

### mishek

L - the length of the rope.

5. Oct 27, 2012

### chikis

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?

Last edited: Oct 27, 2012