2. Gravitational acceleration from a pendulum

1. Introduction and theory

For a simple pendulum

$$ T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow g = 4\pi^2 \frac{L}{T^2}. $$

For a compound pendulum

$$ T = 2\pi \sqrt{\frac{\frac{K^2}{l}+l}{g}} $$

and, hence, a compound pendulum is equivalent to a simple pendulum if

$$ L = \frac{K^2}{l} + l \Rightarrow l^2 - lL + K^2 = 0 $$

which is a quadratic equation with two solutions $l_1$ and $l_2$ where $l_1+l_2=L$ and $l_1l_2=K^2$.

You have to find gravitational acceleration $g$ and radius of gyration $K=\sqrt{l_1l_2}$.

Hole no.	Distance [cm]	Trial
1	10	1
1	10	2
2	20	1
2	20	2
3	30	1
3	30	2
4	40	1
4	40	2
6	60	1
6	60	2
7	70	1
7	70	2
8	80	1
8	80	2
9	90	1
9	90	2

$$ g = \frac{GM}{R^2} $$

Answer the following questions in Discussion.

Why the angle of deflection of the pendulum should not be large?
Why are the periods at 10, 30, 70 and 90 cm similar?
Why do you get two symmetric curves after plotting $T$ as a function of $l$.
If your were periods at 10, 30, 70 and 90 cm were not as similar as expected, discuss why this happened?

Hole no.	Distance [cm]	Trial
1	10	1
1	10	2
2	20	1
2	20	2
3	30	1
3	30	2
4	40	1
4	40	2
6	60	1
6	60	2
7	70	1
7	70	2
8	80	1
8	80	2
9	90	1
9	90	2

Hole no.	Distance [cm]	Trial
1	10	1
1	10	2
2	20	1
2	20	2
3	30	1
3	30	2
4	40	1
4	40	2
6	60	1
6	60	2
7	70	1
7	70	2
8	80	1
8	80	2
9	90	1
9	90	2

Hole no.	Distance [cm]	Trial
1	10	1
1	10	2
2	20	1
2	20	2
3	30	1
3	30	2
4	40	1
4	40	2
6	60	1
6	60	2
7	70	1
7	70	2
8	80	1
8	80	2
9	90	1
9	90	2