| Date | May 2012 | Marks available | 6 | Reference code | 12M.2.SL.TZ2.5 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Describe, Determine, and Draw | Question number | 5 | Adapted from | N/A |
Question
Part 2 Simple harmonic motion and waves
One end of a light spring is attached to a rigid horizontal support.
An object W of mass 0.15 kg is suspended from the other end of the spring. The extension x of the spring is proportional to the force F causing the extension. The force per unit extension of the spring k is 18 Nm–1.
A student pulls W down such that the extension of the spring increases by 0.040 m. The student releases W and as a result W performs simple harmonic motion (SHM).
(i) State what is meant by the expression “W performs SHM”.
(ii) Determine the maximum acceleration of W.
(iii) Determine the period of oscillation of the spring.
(iv) Determine the maximum kinetic energy of W.
A light spring is stretched horizontally and a longitudinal travelling wave is set up in the spring, travelling to the right.
(i) Describe, in terms of the propagation of energy, what is meant by a longitudinal travelling wave.
(ii) The graph shows how the displacement x of one coil C of the spring varies with time t.
The speed of the wave is 3.0 cms–1. Determine the wavelength of the wave.
(iii) Draw, on the graph in (c)(ii), the displacement of a coil of the spring that is 1.8 cm away from C in the direction of travel of the wave, explaining your answer.
Markscheme
(i) the acceleration of (force acting on) W is proportional to its displacement from equilibrium;
and directed towards equilibrium;
(ii) F=(18×0.04=) 0.72N;
acceleration=\(\frac{{0.72}}{{0.15}}\)=4.8 ms-2;
(iii) \(\omega = \sqrt {\frac{a}{x}} \);
=10.95rads-2;
\(T = \left( {\frac{{2\pi }}{\omega } = } \right)\frac{{6.28}}{{10.95}} = 0.57{\rm{s}}\);
(iv) =1.4×10-2(J);
(i) the direction of oscillation of the particles of the medium;
(must see “particles”)
is in the direction of energy propagation;
Accept answer in terms of coils of spring in place of particles of medium.
(ii) frequency=\(\left( {\frac{1}{T} = \frac{1}{{0.80}} = } \right)1.25{\rm{Hz}}\);
wavelength=\(\frac{v}{f} = \frac{{3.0}}{{1.25}} = 2.4{\rm{cm}}\) or 2.4×10-2m;
(iii)
graph: positive cosine; (line must cross axis at 0.2 and 0.6 as shown)
explanation: 1.8 cm is ¾ of a wavelength;
Examiners report
There were a few G2 comments suggesting that this question was off-syllabus as it involved (the G2s claimed) the use of Hooke‟s law. In fact, k was defined as force per unit extension rather than as the spring constant and the whole question was accessible through knowledge based purely on SHM theory.
(i) Many realized that for two marks they were required to state the relationship between acceleration and displacement and give the direction of acceleration.
(ii) and (iii) The determination of the maximum acceleration and the period of oscillation were well done by many at HL. A number of routes were possible for part (iii) and all gained equal credit. Answers were patchier from SL candidates who struggled more than the HL with these tasks.
(iv) SL only Performance was again patchy. Some of the candidates at SL do not seem at ease with the ideas and equations that lie behind SHM theory.
(i) Examiners were disappointed to see that candidates could only rarely give a complete description of a longitudinal travelling wave. Descriptions were vague and rarely made the relative directions of energy propagation and particle displacement clear in an unambiguous way.
(ii) Although many obtained the correct answer, the method used was often unexplained with no obvious link to the graph via a statement of frequency or period. Such non-clarity was penalized.
(iii) Most candidates were unable to make progress with this question and it was frequently left blank. The problem required a recognition that a distance of 1.8 cm corresponded to a ¾λ shift and hence a corresponding shift on the graph.
