You are provided with a retort stand, clamp and boss, a pendulum bob, a piece of thread, and other necessary apparatus. Carry out the fo lowing experiment: ...

Answer Details

Simple pendulum: graph of \(T^{2}\) against \(L\)

The pendulum obeys \(T = 2\pi\sqrt{\dfrac{l}{g}}\), so that \(T^{2} = \dfrac{4\pi^{2}}{g}\,l\). Here the length used for the horizontal axis is \(L = l - 30\ \text{cm}\), and the time \(t\) is measured for 20 complete oscillations, giving \(T = \dfrac{t}{20}\).

Table of readings

S/N	l (cm)	t (s) for 20 osc.	T (s)	T² (s²)	L = l − 30 (cm)
1	130	45.70	2.29	5.22	100
2	110	42.00	2.10	4.41	80
3	90	38.00	1.90	3.61	60
4	70	33.10	1.66	2.74	40
5	50	28.70	1.44	2.06	20

Graph of \(T^{2}\) against \(L\)

Slope and intercept

Reading two widely separated points on the line of best fit, \((L_{1}, T_{1}^{2}) = (100,\ 5.22)\) and \((L_{2}, T_{2}^{2}) = (20,\ 2.06)\):

\[ s = \frac{\Delta T^{2}}{\Delta L} = \frac{5.22 - 2.06}{100 - 20} = \frac{3.16}{80} = 0.039\ \text{s}^{2}\,\text{cm}^{-1} \]

The line cuts the \(T^{2}\) axis at

\[ C = 1.3\ \text{s}^{2}. \]

Evaluation

Taking \(\pi = \dfrac{22}{7}\):

i. \[ k_{1} = \frac{4\pi^{2}}{s} = \frac{4\left(\frac{22}{7}\right)^{2}}{0.039} = \frac{4 \times 9.878}{0.039} = \frac{39.51}{0.039} = 1013\ \text{cm s}^{-2} = 10.13\ \text{m s}^{-2}. \] This is the acceleration of free fall due to gravity, \(g\).

ii. \[ k_{2} = \frac{C}{s} = \frac{1.3}{0.039} = 33.33. \]

Two precautions

1. I displaced the bob through a small angle and released it so that it swung in one vertical plane only, avoiding conical (elliptical) oscillations.
2. I avoided parallax error by reading the metre rule and stop-clock with the line of sight perpendicular to the scale, and counted the oscillations from the central rest position.

(b)

i. The period of oscillation of a body is the time taken for the body to make one complete to-and-fro movement (one full oscillation).

ii. When a body falls freely under its weight alone, the rate of increase of its velocity with time, caused by the Earth's gravitational pull, is the acceleration of free fall due to gravity. It is directed vertically downwards and has a value of about \(10\ \text{m s}^{-2}\) near the Earth's surface.