Show that \({\omega ^2} = \frac{k}{m}\).

One cycle of the variation of displacement with time is shown for two separate mass–spring systems, A and B.

(i) Calculate the frequency of the oscillation of A.

(ii) The springs used in A and B are identical. Show that the mass in A is equal to the mass in B.

The graph shows the variation of the potential energy of A with displacement.

On the axes,

(i) draw a graph to show the variation of kinetic energy with displacement for the mass in A. Label this A.

(ii) sketch a graph to show the variation of kinetic energy with displacement for the mass in B. Label this B.

A 24 Ω resistor is made from a conducting wire.

(i) The diameter of the wire is 0.30 mm and the wire has a resistivity of 1.7\( \times \)10^–8Ωm. Calculate the length of the wire.

(ii) On the axes, draw a graph to show how the resistance of the wire in (d)(i) varies with the diameter of the wire when the length is constant. The data point for the diameter of 0.30 mm has already been plotted for you.

The 24 Ω resistor is covered in an insulating material. Explain the reasons for the differences between the electrical properties of the insulating material and the electrical properties of the wire.

An electric circuit consists of a supply connected to a 24Ω resistor in parallel with a variable resistor of resistance R. The supply has an emf of 12V and an internal resistance of 11Ω.

Power supplies deliver maximum power to an external circuit when the resistance of the external circuit equals the internal resistance of the power supply.

(i) Determine the value of R for this circuit at which maximum power is delivered to the external circuit.

(ii) Calculate the reading on the voltmeter for the value of R you determined in (f)(i).

(iii) Calculate the total power dissipated in the circuit when the maximum power is being delivered to the external circuit.

ma \(=  - \)kx;
\(a =  - \frac{k}{m}x\); (condone lack of negative sign)
\(\left( {{\omega ^2} = \frac{k}{m}} \right)\)

or

implied use of defining equation for simple harmonic motion \(a =  - {\omega ^2}x\);
\(\left( {{\rm{so }}{\omega ^2} = \frac{k}{m}} \right)\)
\(ma =  - kx\) so \(a =  - \left( {\frac{k}{m}} \right)x\);

(i) 0.833 (Hz);

(ii) frequency/period is the same so ω is the same;
k is the same (as springs are identical);
(so m is the same)

(ii) end displacements correct \( \pm \) 0.01m;
maximum lower than 0.16J;
maximum equal to 0.04J \( \pm \) half square;

(i) \(l = \frac{{\pi {d^2}R}}{{4\rho }}\) seen / correct substitution
into equation: \(24 = \frac{{l \times 1.7 \times {{10}^{ - 8}}}}{{\pi  \times {{\left( {0.15 \times {{10}^{ - 3}}} \right)}^2}}}\); } (condone use of r for \(\frac{d}{2}\) in first alternative)

99.7 (m);
Award [2] for bald correct answer.
Award [1 max] if area is incorrectly calculated, answer is 399 m if conversion to radius ignored, ie: allow ECF for second marking point if area is incorrect provided working clear.

(ii) any line showing resistance decreasing with increasing diameter and touching
point;
correct curved shape showing asymptotic behavior on at least one axis;

current/conduction is (related to) flow of charge;
conductors have many electrons free/unbound / electrons are the charge carriers / insulators have few free electrons;
pd/electric field accelerates/exerts force on electrons;
smaller current in insulators as fewer electrons available / larger current in conductors as more electrons available;

(i) use of total resistance = 11Ω; (can be seen in second marking point)
\(\frac{1}{{11}} = \frac{1}{R} + \frac{1}{{24}}\);
20.3(Ω) ;

(ii) as current is same in resistor network and cell and resistance is same, half of emf must appear across resistor network;
6.0 (V);

or

\(I = \frac{{12}}{{\left( {11 + 11} \right)}} = 0.545\left( {\rm{A}} \right)\);
V=(0.545×11=) 6.0 (V);

Other calculations are acceptable.
Award [2] for a bald correct answer.

(iii) use of 22 (ohm) or 11+11 (ohm) seen;
use of \(\frac{{{V^2}}}{R}\) or equivalent;
6.54 (W);
Award [3] for bald correct answer.
Award [2 max] if cell internal resistance ignored, yields 3.27 V.