For particle P,

(i) state how graph 1 shows that its oscillations are not damped.

(ii) calculate the magnitude of its maximum acceleration.

(iii) calculate its speed at t=0.12 s.

(iv) state its direction of motion at t=0.12 s.

Graph 2 shows the variation with position d of the displacement x of particles in the medium at a particular instant of time.

Graph 2

Determine for the longitudinal wave, using graph 1 and graph 2,

(i) the frequency.

(ii) the speed.

Graph 2 – reproduced to assist with answering (c)(i).

(c) The diagram shows the equilibrium positions of six particles in the medium.

(i) On the diagram above, draw crosses to indicate the positions of these six particles at the instant of time when the displacement is given by graph 2.

(ii) On the diagram above, label with the letter C a particle that is at the centre of a compression.

(i) the amplitude is constant;

(ii) period is 0.20s;

\({a_{\max }} = \left( {{{\left[ {\frac{{2\pi }}{T}} \right]}^2}{x_0} = {{31.4}^2} \times 2.0 \times {{10}^{ - 2}}} \right) = 19.7 \approx 20{\rm{m}}{{\rm{s}}^{ - 2}}\)

Award [2] for correct bald answer and ignore any negative signs in answer.

(iii) displacement at t = 0.12cm is (−)1.62cm;
\(v\left( { = \frac{{2\pi }}{T}\sqrt {{x_0} - {x^2}} } \right) = 31.4\sqrt {\left( {2.0 \times {{10}^{ - 2}}} \right)^2 - {{\left( {1.62 \times {{10}^{ - 2}}} \right)}^2}}  = 0.37{\rm{m}}{{\rm{s}}^{ - 1}}\);
Accept displacement in range 1.60 to 1.70 cm for an answer in range 0.33ms⁻¹ to 0.38ms⁻¹.

or

\({v_0} = \frac{{2\pi }}{T}{x_0} = 0.628{\rm{m}}{{\rm{s}}^{ - 1}}\);
\(\left| {\left. v \right|} \right. = \left( {\left| {\left. { - {v_0}\sin \left[ {\frac{{2\pi }}{T}t} \right]} \right|} \right. \Rightarrow \left| {\left. v \right|} \right. = \left| {\left. { - 0.628\sin \left[ {31.4 \times 0.12} \right]} \right|} \right. = \left| {\left. {0.37} \right|} \right.} \right) = 0.37{\rm{m}}{{\rm{s}}^{ - 1}}\);

or

drawing a tangent at 0.12s;
measurement of slope of tangent;
Accept answer in range 0.33ms⁻¹ to 0.38ms⁻¹ .

(i) use of \(f = \frac{1}{T}\);
and so \(f\left( { = \frac{1}{{0.20}}} \right) = 5.0{\rm{Hz}}\);

(ii) wavelength is 16cm;
and so speed is v(=fλ=5.0×0.16)=0.80ms⁻¹;

(i) points at 0, 8 and 16 cm stay in the same place;
points at 4 and 20 cm move 2 cm to the right;
point at 12 cm moves 2 cm to the left;

(ii) the point at 8 cm;