2022 H2 Physics Paper 2 Q1 — A Spring Constant With Its Uncertainty, and a v

The model, live

Value / pathBoundsDamped

Part (a)

4 marks

Ten oscillations take $(7.2 \pm 0.2)\,\text{s}$ and $m = 120\,\text{g} \pm 1\%$ . Using $T = 2\pi\sqrt{m/k}$ , find $k$ with its actual uncertainty, to an appropriate number of significant figures.

Approach

Rearrange for $k$ , then combine fractional uncertainties: $k \propto m\,T^{-2}$ , so the fractional uncertainty in $T$ counts twice. The uncertainty in the time for ten oscillations has the same fractional size as the uncertainty in one period. Round the uncertainty to one significant figure, then round $k$ to match it.

T = \frac{7.2}{10} = 0.72\,\text{s} \qquad k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2(0.120)}{(0.72)^2} = 9.14\,\text{N m}^{-1}

\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T} = 0.01 + 2\left(\frac{0.2}{7.2}\right) = 0.0656

\Delta k = 0.0656 \times 9.14 = 0.60\,\text{N m}^{-1}

To check your answer:

Take the extremes. The largest $k$ comes from the heaviest mass and the shortest time, $4\pi^2(0.1212)/(0.700)^2 = 9.77$ ; the smallest from the lightest mass and the longest time, $4\pi^2(0.1188)/(0.740)^2 = 8.56$ . Half the spread is $0.60$ , the same as the uncertainty calculated.

$k = (9.1 \pm 0.6)\,\text{N m}^{-1}$

Watch

Writing $9.14 \pm 0.6$ loses the mark for significant figures. The uncertainty is in the first decimal place, so $k$ stops there too.

Part (b)(i)

3 marks

Use the velocity–displacement graph to find the maximum acceleration of the mass.

Approach

The ellipse gives two numbers: where it crosses the $x$ axis is the amplitude, and where it crosses the $v$ axis is the maximum speed. Their ratio is $\omega$ , since $v_0 = \omega x_0$ . Then $a_\text{max} = \omega^2 x_0$ .

From the graph, $x_0 = 0.16\,\text{m}$ and $v_0 = 1.4\,\text{m s}^{-1}$ :

\omega = \frac{v_0}{x_0} = \frac{1.4}{0.16} = 8.75\,\text{rad s}^{-1}

a_\text{max} = \omega^2 x_0 = (8.75)^2(0.16) = 12\,\text{m s}^{-2}

To check your answer:

From part (a), $\omega^2 = k/m = 9.14/0.120 = 76$ , so $a_\text{max} = 76 \times 0.16 = 12\,\text{m s}^{-2}$ . Equivalently $a_\text{max} = v_0^2/x_0 = 1.4^2/0.16 = 12\,\text{m s}^{-2}$ .

$a = 12\,\text{m s}^{-2}$

Part (b)(ii)

2 marks

The oscillations are now lightly damped. Draw the v–x graph for the first complete cycle.

A curve starting at the release point ( $x = 0.16\,\text{m}$ , $v = 0$ ), going round in the same sense as the ellipse but spiralling inwards: each crossing of the $v$ axis is a little lower than before, and after one cycle it reaches $v = 0$ at a displacement a little less than $0.16\,\text{m}$ . It does not close.

Note

Light damping hardly changes the period, so the spiral keeps the ellipse's shape; it just shrinks a little each half-cycle as energy is lost to the air.

What this question was really testing

Combining uncertainties. Powers multiply fractional uncertainties: $T^2$ doubles $\Delta T/T$ .
Reporting a result. One significant figure in the uncertainty, and the value rounded to the same decimal place.
Reading a v–x graph. Intercepts are $x_0$ and $v_0$ ; their ratio is $\omega$ .
Damping as lost energy. A closed loop becomes an inward spiral.

A spring constant to the right number of figures, and an ellipse that spirals in

Part (a)

Part (b)(i)

Part (b)(ii)

What this question was really testing

This is how we mark, and how we teach.