A particle P moves with simple harmonic motion. State, with reference to the motion of P, what is meant by simple harmonic motion.

Use the graph opposite to determine for the motion of P the

(i) period.

(ii) amplitude.

(iii) displacement of P from equilibrium at t=0.2s.

The particle P in (b) is a particle in medium M₁ through which a transverse wave is travelling.

(i) Describe, in terms of energy propagation, what is meant by a transverse wave.

(ii) The speed of the wave through the medium is 0.40ms^–1. Calculate, using your answer to (b)(i), the wavelength of the wave.

(iii) The wave travels into another medium M₂. The refractive index of M₂ relative to M₁ is 1.8. Calculate the wavelength of the wave in M₂.

Outline, with reference to the graph and to Ohm’s law, whether or not each component is ohmic.

Components X and Y are connected in parallel. The parallel combination is then connected in series with a variable resistor R and a cell of emf 8.0V and negligible internal resistance.

The resistance of R is adjusted until the currents in X and Y are equal.

(i) Using the graph, calculate the resistance of the parallel combination of X and Y.

(ii) Using your answer to (e)(i), determine the resistance of R.

(iii) Determine the power delivered by the cell to the circuit.

the acceleration (of a particle/P) is (directly) proportional to displacement;
and is directed towards equilibrium/in the opposite direction to displacement;
Do not accept “directed towards the centre”.

(i) 0.30 s;

(ii) max velocity=0.74(±0.02)ms⁻¹;
recognize max velocity=ωx₀;
\(\omega  = \left( {\frac{{2\pi }}{T} = \frac{{2\pi }}{{0.30}} = } \right)20.9{\rm{rad }}{{\rm{s}}^{ - 1}}\);
\({x_0} = \left( {\frac{{0.74}}{{20.9}} = } \right)3.5\left( { \pm 2.0} \right) \times {10^{ - 2}}{\rm{m}}\);

or

identifies displacement with area;
uses one quarter of a cycle;
answer in the range of 30 to 40 mm;
answer in the range of 33 to 37 mm;

(iii) \(v = 0.64\left( { \pm 2.0} \right){\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}\);
use \(v = \omega \sqrt {\left( {{x_0}^2 - {x^2}} \right)} \) to get x=1.7(±0.2)×10^-2m

or

recognition that x=x₀cosωt;
\(x\left( { = 35\cos \left[ {\frac{{2\pi }}{{0.3}} \times 0.2} \right]} \right) = 17.5{\rm{mm}}\);

(i) the direction of energy propagation is at right angles to the motion of the particles/atoms/molecules in the medium;

(ii) \(\begin{array}{l}
\lambda  = \frac{v}{f} = vT;\\
 = \left( {0.40 \times 0.3 = } \right)0.12{\rm{m;}}
\end{array}\)

(iii) \(n/1.8 = \frac{{{v_1}}}{{{v_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\);
to give \({{\lambda _2}}\)=0.067m;

X: graph is a straight line and through the origin / resistance is constant;
so because V∝ I it is ohmic;

Y: not ohmic because graph is not straight/is curved / resistance is not constant;
Award [3] for an answer where resistance values are calculated to show constancy or otherwise.

(i) read-off of intersection of lines X and Y [4.0,6.0] / reference to 4.0V and 6.0mA; { (allow power of 10 error)
\({R_X} = {R_Y} = \frac{{6.0}}{{4.0 \times {{10}^{ - 3}}}} = 1.5 \times {10^3}\Omega \);
resistance of combination=750Ω;

(ii) use the idea of potential divider \(\frac{R}{{750}} = \frac{{2.0}}{{6.0}}\);
R=250Ω;

or

current=8mA;
\(R = \frac{{2.0}}{{0.008}} = 250\left( \Omega  \right)\);

(iii) total resistance=1000Ω;
total current=8.0×10^-3A or pd=8.0V;
total power=(8.0×8.0×10^-3=)64mW;