Question 1 Report
You have been provided with a metre rule, a clamp, and a set of masses.
(b)i. Explain simple harmonic motion.
ii. Define period and frequency, with respect to a simple harmonic motion.
Precautions:
(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.
This is a loaded-cantilever oscillation experiment. The metre rule projecting 90 cm from the bench behaves like a spring; when a mass \(M\) fixed at its free end is deflected and released, it performs vertical simple harmonic motion. For each mass you time 10 complete oscillations, find the period \(T=\dfrac{t}{10}\), and compute \(T^2\). Theory gives \(T^2=\dfrac{4\pi^2}{k}M + C\), so a graph of \(T^2\) against \(M\) is a straight line whose slope allows \(k\) to be found.
Readings and table of values
| M (g) | t (s) for 10 oscillations | T = t/10 (s) | T2 (s2) |
|---|---|---|---|
| 50 | 3.16 | 0.316 | 0.100 |
| 100 | 3.87 | 0.387 | 0.150 |
| 150 | 4.47 | 0.447 | 0.200 |
| 200 | 5.00 | 0.500 | 0.250 |
| 250 | 5.48 | 0.548 | 0.300 |
Graph of T2 against M
Slope and intercept
Using two convenient points on the line, \((M_1,T_1^2)=(50,\,0.100)\) and \((M_2,T_2^2)=(250,\,0.300)\):
\[ s=\frac{\Delta T^2}{\Delta M}=\frac{0.300-0.100}{250-50}=\frac{0.200}{200}=1.0\times10^{-3}\ \text{s}^2\,\text{g}^{-1} \]The line meets the vertical axis where \(M=0\), giving the intercept
\[ C = 0.05\ \text{s}^2 \]Evaluation of k
From \(T^2=\dfrac{4\pi^2}{k}M + C\), the slope is \(s=\dfrac{4\pi^2}{k}\), so \(k=\dfrac{4\pi^2}{s}\). With \(\pi=\dfrac{22}{7}\):
\[ k=\frac{4\pi^2}{s}=\frac{4\times\left(\frac{22}{7}\right)^2}{1.0\times10^{-3}}=\frac{4\times 9.878}{1.0\times10^{-3}}=\frac{39.51}{1.0\times10^{-3}}\approx 3.95\times10^{4} \]Period when M = 180 g
Reading up from \(M=180\ \text{g}\) to the line and across to the vertical axis gives \(T^2=0.23\ \text{s}^2\), so
\[ T=\sqrt{0.23}=0.48\ \text{s} \]Two precautions
(b)(i) Simple harmonic motion
Simple harmonic motion is the motion of a body in which its acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point.
(b)(ii) Period and frequency
The period is the time taken by the oscillating body to make one complete oscillation. The frequency is the number of complete oscillations made in one second, related to the period by \(f=\dfrac{1}{T}\).
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