**APPARATUS:**
Retort stand, Split cork, Optical pin, Perforated Meter rule, Stopwatch.
**PROCEDURE:**
I. Hang the metre rule at the point mark, L = 90.0 cm as shown in the diagram above.
II. Displace the metre rule slightly and record two sets of time (t₁ and t₂) for 10 complete oscillations.
III. Obtain the average time for the oscillations.
IV. Repeat the procedures in **II** and **III** for different values of L: 85.0 cm, 80.0 cm, 75.0 cm and 70.0 cm respectively.
V. Tabulate your readings as shown below:
| L (cm) | t₁ (s) | t₂ (s) | t = (t₁+t₂)/2 (s) | T (s) | T² (s²) |
|--------|--------|--------|-------------------|-------|---------|
| 90.0 | | | | | |
| 85.0 | | | | | |
| 80.0 | | | | | |
| 75.0 | | | | | |
| 70.0 | | | | | |
VI. Plot a graph of T² against L.
VII. Determine the slope, S of the graph.
VIII. Given the period of oscillation:
$$T = 2\pi \sqrt{\frac{I}{Mgd}}$$
where I is the moment of inertia of the metre rule given as:
$$I = \frac{1}{3}ML^2$$
and d = L/2. Using your graph, calculate the value of the acceleration of gravity, **g**.
IX. Define Moment of Inertia, and state its SI unit.
# Perforated Metre Rule Oscillation Experiment — Full Solution
## BACKGROUND THEORY
Given:
- $T = 2\pi \sqrt{\dfrac{I}{Mgd}}$
- $I = \dfrac{1}{3}ML^2$
- $d = \dfrac{L}{2}$
**Simplify T:**
Substitute I and d into the formula:
$$T = 2\pi \sqrt{\frac{\frac{1}{3}ML^2}{Mg \cdot \frac{L}{2}}}$$
$$T = 2\pi \sqrt{\frac{\frac{1}{3}L^2}{\frac{gL}{2}}}$$
$$T = 2\pi \sqrt{\frac{1}{3}L^2 \times \frac{2}{gL}}$$
$$T = 2\pi \sqrt{\frac{2L}{3g}}$$
Squaring both sides:
$$\boxed{T^2 = \frac{8\pi^2}{3g} \cdot L}$$
This is of the form **T² = SL**, where slope:
$$S = \frac{8\pi^2}{3g}$$
Therefore:
$$\boxed{g = \frac{8\pi^2}{3S}}$$
---
## STEP 1 — Tabulate Readings (Typical Expected Values)
For a metre rule oscillating as a physical pendulum, use the formula $T = 2\pi\sqrt{\frac{2L}{3g}}$ with g = 9.8 m/s² to calculate expected T values.
> **Note:** Convert L to metres first.
| L (cm) | L (m) | T (s) calculated | T² (s²) |
|--------|-------|-----------------|---------|
| 90.0 | 0.90 | 1.554 | 2.415 |
| 85.0 | 0.85 | 1.510 | 2.280 |
| 80.0 | 0.80 | 1.464 | 2.143 |
| 75.0 | 0.75 | 1.417 | 2.008 |
| 70.0 | 0.70 | 1.369 | 1.874 |
**Sample calculation for L = 90.0 cm:**
$$T = 2\pi\sqrt{\frac{2 \times 0.90}{3 \times 9.8}} = 2\pi\sqrt{\frac{1.80}{29.4}} = 2\pi\sqrt{0.06122} = 2\pi \times 0.2474 = 1.554 \text{ s}$$
$$T^2 = (1.554)^2 = 2.415 \text{ s}^2$$
## STEP 2 — t₁, t₂ Values (for 10 oscillations)
Multiply each T by 10 to get time for 10 oscillations, then simulate t₁ and t₂ (slightly different as in real experiment):
| L (cm) | t₁ (s) | t₂ (s) | t = (t₁+t₂)/2 | T = t/10 (s) | T² (s²) |
|--------|--------|--------|---------------|-------------|---------|
| 90.0 | 15.6 | 15.5 | 15.55 | 1.555 | 2.418 |
| 85.0 | 15.1 | 15.1 | 15.10 | 1.510 | 2.280 |
| 80.0 | 14.6 | 14.7 | 14.65 | 1.465 | 2.147 |
| 75.0 | 14.2 | 14.1 | 14.15 | 1.415 | 2.002 |
| 70.0 | 13.7 | 13.6 | 13.65 | 1.365 | 1.863 |
---
## STEP 3 — Plot Graph of T² against L
- **Y-axis:** T² (s²)
- **X-axis:** L (m)
- Plot 5 points and draw a best-fit straight line through the origin
**Points to plot:**
| L (m) | T² (s²) |
|-------|---------|
| 0.70 | 1.863 |
| 0.75 | 2.002 |
| 0.80 | 2.147 |
| 0.85 | 2.280 |
| 0.90 | 2.418 |
The graph is a **straight line through the origin** ✓
## STEP 4 — Determine the Slope S
Pick two well-separated points on the best-fit line:
Use **(0.70, 1.863)** and **(0.90, 2.418)**:
$$S = \frac{\Delta T^2}{\Delta L} = \frac{2.418 - 1.863}{0.90 - 0.70} = \frac{0.555}{0.20}$$
$$\boxed{S = 2.775 \text{ s}^2\text{m}^{-1}}$$
## STEP 5 — Calculate g
$$g = \frac{8\pi^2}{3S}$$
$$g = \frac{8 \times (3.142)^2}{3 \times 2.775}$$
$$g = \frac{8 \times 9.870}{8.325}$$
$$g = \frac{78.957}{8.325}$$
$$\boxed{g \approx 9.484 \approx 9.5 \text{ ms}^{-2}}$$
This is close to the standard value of **9.8 ms⁻²** ✓ (small deviation is acceptable in experiments)
## STEP 6 — Question IX: Moment of Inertia
**Definition:**
> The Moment of Inertia of a body is the sum of the products of each particle's mass and the square of its perpendicular distance from the axis of rotation.
Mathematically: $I = \sum mr^2$
**SI Unit:** **kg·m²**
## SUMMARY TABLE
| Step | What you do | Result |
|------|------------|--------|
| Simplify T | Substitute I and d | $T^2 = \frac{8\pi^2}{3g}L$ |
| Table | Compute t, T, T² | See table above |
| Graph | Plot T² vs L | Straight line through origin |
| Slope | Rise ÷ Run | S = 2.775 s²m⁻¹ |
| Find g | $g = \frac{8\pi^2}{3S}$ | g ≈ 9.5 ms⁻² |
| Moment of Inertia | Define + unit | kg·m² |