Young modulus experiment
Materials - OCR A-Level Physics
Goal of the experiment
Measure the Young modulusThe ratio of stress to strain for a material in the elastic region. A measure of stiffness. Measured in pascals (Pa). $E$ of a metal wire by stretching it with known loads, recording extension and original length, and plotting stress against strain. The gradient of the linear region equals $E$.
$$E = \frac{\sigma}{\varepsilon} = \frac{FL}{A \, \Delta L}$$
Diagram pending
Young modulus apparatus. A long wire clamped at one end, passing over a pulley at the bench edge to a hanging mass holder. A fixed reference marker on the wire is read against a metre rule. A second identical reference wire alongside compensates for thermal expansion.
Will be replaced with a GeoGebra SVG in stream 2.
- Use a long, thin wire (at least 2 m) clamped at one end with a hanger and masses at the other. A long wire gives a larger extension for the same strain, cutting percentage uncertainty in $\Delta L$.
- Measure the original length $L$ from the clamp to the reference marker with a metre rule. Measure the diameter $d$ with a micrometer screw gaugeA precision instrument for measuring small thicknesses or diameters to about $\pm 0.01 \text{ mm}$. at three points along the wire and in two perpendicular directions, then take the mean. Compute $A = \pi (d/2)^2$.
- Add masses in steps of about 100 g. After each mass, measure the new extension $\Delta L$ with a travelling microscopeAn instrument for measuring small displacements by sliding a crosshair on a calibrated scale. or a marker viewed against a fixed scale.
- Plot stress $\sigma = F/A$ on the $y$-axis against strain $\varepsilon = \Delta L / L$ on the $x$-axis. The gradient of the linear region is the Young modulus.
- Use a reference wire alongside the test wire to cancel thermal expansion of the lab. Stop loading well before the elastic limit so the formula stays valid.
Worked Example [4 marks]
A steel wire of original length $2.5 \text{ m}$ and diameter $0.50 \text{ mm}$ is stretched by a force of $80 \text{ N}$. The extension is $1.2 \text{ mm}$. Calculate the Young modulus of the steel.
Show Solution
1
Radius $r = d/2 = 0.25 \times 10^{-3} \text{ m}$. Area $A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7} \text{ m}^2$.
[1]2
Stress $\sigma = F/A = 80 / (1.96 \times 10^{-7}) = 4.08 \times 10^{8} \text{ Pa}$.
[1]3
Strain $\varepsilon = \Delta L / L = (1.2 \times 10^{-3}) / 2.5 = 4.8 \times 10^{-4}$.
[1]4
$E = \sigma / \varepsilon = (4.08 \times 10^{8}) / (4.8 \times 10^{-4}) = 8.5 \times 10^{11} \text{ Pa} \approx 200 \text{ GPa}$.
[1]Answer
$E \approx 2.0 \times 10^{11} \text{ Pa}$ (about $200 \text{ GPa}$, consistent with steel).
Common Mistake
HIGH
Wrong: Using the wire diameter as the radius when computing area, giving $A$ four times too big and $E$ four times too small. Or mixing mm with m in the same calculation.
Right: $r = d/2$, then $A = \pi r^2$. Convert every length to metres before substituting: $0.50 \text{ mm} = 5.0 \times 10^{-4} \text{ m}$, $1.2 \text{ mm} = 1.2 \times 10^{-3} \text{ m}$.
Right: $r = d/2$, then $A = \pi r^2$. Convert every length to metres before substituting: $0.50 \text{ mm} = 5.0 \times 10^{-4} \text{ m}$, $1.2 \text{ mm} = 1.2 \times 10^{-3} \text{ m}$.
Examiner Tips and Tricks
- Stock answer to 'why a long thin wire?': long $\Rightarrow$ larger extension for the same strain $\Rightarrow$ smaller percentage uncertainty in $\Delta L$. Thin $\Rightarrow$ higher stress for the same load $\Rightarrow$ the wire reaches the linear region without huge masses.
- Always state safety: clamp the masses over a tray or sand bag in case the wire snaps.