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: ...
Question 1 Report
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:
- Set up the apparatus as shown in the diagram above:
- measure and record the distance = 130cm from the centre of the bob to the point of suspension of the pendulum
- displace the pendulum through a small angle and release. Allow the pendulum to oscillate freely;
- determine the time t for 20 complete oscillations;
- also, determine the period \(T\) of the oscillations
- evaluate \(T^{2}\) and \(L = -30\);
- repeat the procedure for four other values of \(l = 110, 90, 70,\) and \(50\ \text{cm}\)
- in each case, determine \(t\) and evaluate \(T, T^{2}\) and \(L\). Tabulate your readings.
- plot of a graph of \(T^{2}\) on the vertical axis against \(L\) on the horizontal axis, starting both axes from the origin \((0,0)\)
- determine the slope, \(s\), of the graph. Also determine the intercept, \(C\), of the graph on the \(T^{2}\) axis;
- evaluate: i. \(k_{1} = \frac{4\pi^{2}}{s_{1}}\),
ii. \(k_{2} = \frac{c}{8}\) [Take \(\pi = \frac{22}{7}\)] - state two precautions taken to ensure accurate results.
(b)i. What is meant by the period of oscillation of an oscillating body?
i. Explain the acceleration of free fall due to gravity.
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
- I displaced the bob through a small angle and released it so that it swung in one vertical plane only, avoiding conical (elliptical) oscillations.
- 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.
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