TEST OF PRACTICAL KNOWLEDGE QUESTION
You are provided with two retort stands, two-metre rules, pieces of thread and other necessary apparatus.
i. Set up the apparatus as illustrated above ensuring the strings are permanently 10cm from either end of the rule.
ii. Measure and record the length L = 80 cm of the two strings.
iii. Hold both ends of the rule and displace the rule slightly, then release so that it oscillates about a vertical axis through its centre.
iv. Determine and record the time t for 10 complete oscillations.
v. Determine the period T of oscillations.
vi. Evaluate log T and L.
vii. Repeat the procedure for four other values of L= 70 cm, 60 cm, 50 cm, and 40 cm
viii. Tabulate your readings.
ix. Plot a graph with log T on the vertical axis and log L on the horizontal axis.
x. Determine the slope, s, and the intercept, c on the vertical axis.
xi. State two precautions taken to ensure accurate results.
(b)i. Define simple harmonic motion.
ii. Determine the value of L corresponding to t= 12 s from the graph in 1.
Practical: torsional oscillation of a suspended metre rule
For each length \( L \), the period is found from the timing:
\[ T = \frac{t}{10} \]
where \( t \) is the time for 10 complete oscillations. You then compute \( \log T \) and \( \log L \) for the five values (\( L = 80, 70, 60, 50, 40\,\text{cm} \)) and tabulate them.
Table (headings expected)
| L /cm | t /s | T /s | log T | log L |
|---|
| 80 | ... | t/10 | ... | 1.903 |
| 70 | ... | t/10 | ... | 1.845 |
| 60 | ... | t/10 | ... | 1.778 |
| 50 | ... | t/10 | ... | 1.699 |
| 40 | ... | t/10 | ... | 1.602 |
Graph: plotting \( \log T \) (vertical) against \( \log L \) (horizontal) gives a straight line, since \( T \propto L^{s} \) means \( \log T = s\log L + c \). The slope s is read from \( \dfrac{\Delta(\log T)}{\Delta(\log L)} \) and the intercept c is the value of \( \log T \) where the line cuts the vertical axis.
Two precautions
- Avoid parallax error when reading the metre rule and when starting/stopping the stopwatch.
- Ensure the rule twists about a vertical axis through its centre (a genuine torsional oscillation) with only a small angular displacement, and count the oscillations from a fixed reference mark.
(b)(i) Simple harmonic motion
SHM is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point: \( a = -\omega^2 x \).
(b)(ii) For \( t = 12\,\text{s} \), the period is \( T = 12/10 = 1.2\,\text{s} \), so \( \log T = \log 1.2 = 0.079 \). Reading across from \( \log T = 0.079 \) on your line to the horizontal axis gives \( \log L \); then \( L = 10^{\log L} \). The exact value must be taken from your own plotted graph.
Practical: torsional oscillation of a suspended metre rule
For each length \( L \), the period is found from the timing:
\[ T = \frac{t}{10} \]
where \( t \) is the time for 10 complete oscillations. You then compute \( \log T \) and \( \log L \) for the five values (\( L = 80, 70, 60, 50, 40\,\text{cm} \)) and tabulate them.
Table (headings expected)
| L /cm | t /s | T /s | log T | log L |
|---|
| 80 | ... | t/10 | ... | 1.903 |
| 70 | ... | t/10 | ... | 1.845 |
| 60 | ... | t/10 | ... | 1.778 |
| 50 | ... | t/10 | ... | 1.699 |
| 40 | ... | t/10 | ... | 1.602 |
Graph: plotting \( \log T \) (vertical) against \( \log L \) (horizontal) gives a straight line, since \( T \propto L^{s} \) means \( \log T = s\log L + c \). The slope s is read from \( \dfrac{\Delta(\log T)}{\Delta(\log L)} \) and the intercept c is the value of \( \log T \) where the line cuts the vertical axis.
Two precautions
- Avoid parallax error when reading the metre rule and when starting/stopping the stopwatch.
- Ensure the rule twists about a vertical axis through its centre (a genuine torsional oscillation) with only a small angular displacement, and count the oscillations from a fixed reference mark.
(b)(i) Simple harmonic motion
SHM is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that fixed point: \( a = -\omega^2 x \).
(b)(ii) For \( t = 12\,\text{s} \), the period is \( T = 12/10 = 1.2\,\text{s} \), so \( \log T = \log 1.2 = 0.079 \). Reading across from \( \log T = 0.079 \) on your line to the horizontal axis gives \( \log L \); then \( L = 10^{\log L} \). The exact value must be taken from your own plotted graph.
