| Date | May 2011 | Marks available | 9 | Reference code | 11M.2.SL.TZ2.4 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine, Draw, and Show that | Question number | 4 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about simple harmonic motion (SHM) and a wave in a string. Part 2 is about the unified atomic mass unit and a nuclear reaction.
Part 1 Simple harmonic motion and a wave in a string
A liquid is contained in a U-tube.
The pressure on the liquid in one side of the tube is increased so that the liquid is displaced as shown in diagram 2. When the pressure is suddenly released the liquid oscillates. The damping of the oscillations is small.
(i) Describe what is meant by damping.
(ii) The displacement of the liquid surface from its equilibrium position is x. The acceleration a of the liquid in the tube is given by the expression
\[a = - \frac{{2g}}{l}x\]
where g is the acceleration of free fall and l is the total length of the liquid column. The total length of the liquid column in the tube is 0.32m. Determine the period of oscillation.
A wave is travelling along a string. The string can be modelled as a single line of particles and each particle executes simple harmonic motion. The period of oscillation of the particles is 0.80s.
The graph shows the displacement y of part of the string at time t=0. The distance along the string is d.
(iv) On the graph opposite, label with the letter X the position of particle P at t=0.40 s.
Markscheme
the maximum displacement of the system from equilibrium/from centre of motion / OWTTE;
(i) the amplitude of the oscillations/(total) energy decreases (with time); because a force always opposes direction of motion/there is a resistive force/ there is a friction force;
Do not allow bald “friction”.
(ii) \(\omega = \sqrt {\frac{{2g}}{l}} \);
\(T = 2\pi \sqrt {\frac{{0.32}}{{2 \times 9.81}}} \);
=0.80s;
(ii) y0 = 0.050(m) and y = 0.030(m);
\(\omega = \left( {\frac{{2\pi }}{{0.80}} = } \right)7.85\left( {{\rm{rad}}{{\rm{s}}^{ - 1}}} \right)\);
\(v = 7.85\sqrt {{{\left[ {0.05} \right]}^2} - {{\left[ {0.03} \right]}^2}} \);
=0.31ms−1 ; (allow working in cm to give 31 cms-1);
(iii) λ=4.0m;
recognition that \(f = \frac{1}{{0.80}}\left( { = 1.25} \right)\);
(fλ=)v=1.25×4.0;
(=5.0 ms−1)
(iv) y=-3.0 cm, d=0.6 m;
