You have been provided with a metre rule, a clamp, and a set of masses.
- Clamp the metre rule to the edge of the bench such that 90cm of the rule projects from the edge as shown in the diagram above. Ensure that the rule is capable of performing oscillatory motion.
- Fix a mass M = 50g at the free end of the rule.
- Deflect the rule slightly such that it performs vertical oscillation.
- Determine the time t for 10 complete oscillations.
- Calculate the period T of the oscillations and evaluate T\(^{2}\)
- Repeat the procedure for four other values of M = 100, 150, 200, and 250g. In each case determine and record the corresponding values of t, T, and T\(^{2}\). Tabulate your readings.
- Plot a graph of T\(^{2}\) on the vertical axis against M on the horizontal axis, starting both axes from the origin (0,0).
- Determine the slopes, of the graph and its intercept C on the vertical axis.
- Evaluate k = 4\(\pi\)/s. [Take \(\pi\) = \(\frac{22}{7}\)].
- From your graph, determine the period T, when M= 180g.
- State two precautions taken to ensure accurate results.
(b)i. Explain simple harmonic motion.
ii. Define period and frequency, with respect to a simple harmonic motion.
Precautions:
- l ensured that the metre rule was firmly clamped
- Readings were repeated
- Parallax was avoided when readings on the stopwatch/clock were taken.
- zero error was noted and corrected on the stopwatch/clock.
(b)i. Simple harmonic motion is a motion in which the acceleration is proportional to the displacement from a fixed point and is directed towards the point.
ii. Period is the time taken by an oscillatory body to make one complete oscillation.
Frequency: is the number of complete oscillations performed in one second.
Explanation
Table of values (observation
| Mass (M)g | Time (t) sec. | T | T\(^{2}\) |
| 1 | 3.5 | 0.35 | 0.123 |
| 2 | 4.2 | 0.42 | 0.176 |
| 3 | 4.9 | 0.49 | 0.240 |
| 4 | 5.5 | 0.55 | 0.303 |
| 5 | 6.0 | 0.60 | 0.360 |
Slope (s) = \(\frac{y}{x} = \frac{\bigtriangleup {T}^{2}}{\bigtriangleup{M}} = \frac{0.26}{220}\) = 0.001
Intercept (c) on the vertical axis = 0.65
Evaluate, K = \(\frac{4 \pi}{S} = \frac{4 \times {\frac{22}{7}}}{0.001}\)
\(\frac{12.57}{0.0001}\) = 12571
From the graph the value of period T, where M = 180g is;
= \(\sqrt 0.275\) = 0.524, i.e T = 0.524