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Date November 2013 Marks available 2 Reference code 13N.2.SL.TZ0.5
Level Standard level Paper Paper 2 Time zone Time zone 0
Command term State Question number 5 Adapted from N/A

Question

This question is in two parts. Part 1 is about simple harmonic motion (SHM) and waves. Part 2 is about voltage–current (VI) characteristics.

Part 1 Simple harmonic motion (SHM) and waves

Part 2 Voltage–current (VI) characteristics

The graph shows the voltage–current (VI) characteristics, at constant temperature, of two electrical components X and Y.

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

[2]
a.

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.

[7]
b.

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.

[5]
c.

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

[3]
d.

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.

[8]
e.

Markscheme

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”.

a.

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

b.

(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;

c.

X: graph is a straight line and through the origin / resistance is constant;
so because VI 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.

d.

(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;

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Core » Topic 4: Waves » 4.1 – Oscillations
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