| Date | May 2016 | Marks available | 4 | Reference code | 16M.2.HL.TZ0.4 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | Calculate and Show that | Question number | 4 | Adapted from | N/A |
Question
A longitudinal wave is travelling in a medium from left to right. The graph shows the variation with distance x of the displacement y of the particles in the medium. The solid line and the dotted line show the displacement at t=0 and t=0.882 ms, respectively.
The period of the wave is greater than 0.882 ms. A displacement to the right of the equilibrium position is positive.
(i) Calculate the speed of this wave.
(ii) Show that the angular frequency of oscillations of a particle in the medium is ω=1.3×103rads−1.
One particle in the medium has its equilibrium position at x=1.00 m.
(i) State and explain the direction of motion for this particle at t=0.
(ii) Show that the speed of this particle at t=0.882 ms is 4.9ms−1.
The travelling wave in (b) is directed at the open end of a tube of length 1.20 m. The other end of the tube is closed.
(i) Describe how a standing wave is formed.
(ii) Demonstrate, using a calculation, that a standing wave will be established in this tube.
Markscheme
(i)
ALTERNATIVE 1
«distance travelled by wave =» 0.30 m
\(v = \ll \frac{{{\rm{distance}}}}{{{\rm{time}}}} = \gg 340{\rm{m}}{{\rm{s}}^{ - 1}}\)
ALTERNATIVE 2
evaluates \(T = \frac{{{\rm{0.882}} \times {{10}^{ - 3}} \times 1.6}}{{{\rm{0.3}}}}\)«=4.7ms» to give f = 210 or 212 Hz
uses λ=1.6 m with v=fλ to give 340ms–1
(ii)
ALTERNATIVE 1
λ=1.60m
\(\omega = \ll 2\pi f = \gg 2\pi \times \frac{{340}}{{1.60}} = 1.3 \times {10^3}\) or 1.34×103rads–1
ALTERNATIVE 2
«0.882 ms is \(\frac{{0.3}}{{1.6}}\) of cycle so whole cycle is» \(\frac{{2\pi \times 3}}{{16 \times 0.882 \times {{10}^{ - 3}}}}\)
1.35×103rads–1
Allow ECF from (b)(i).
(i)
the displacement of the particle decreases OR «on the graph» displacement is going in a negative direction OR on the graph the particle goes down OR on the graph displacement moves towards equilibrium/0
to the left
Do not allow “moving downwards”.
(ii)
y=–1.5mm
\(v = 2\pi \times 212 \times \sqrt {{{\left( {4.0 \times {{10}^{ - 3}}} \right)}^2} - {{\left( {1.5 \times {{10}^{ - 3}}} \right)}^2}} \)
«v=4.939≈4.9ms-1»
Allow ECF from (b)(ii).
Do not allow \(\frac{{4.3{\rm{mm}}}}{{0.882{\rm{ms}}}} = 4.87{\rm{m}}{{\rm{s}}^{ - 1}}\).
(i)
the superposition/interference of two oppositely moving/reflected «identical travelling» waves
(ii)
the allowed wavelengths in the tube are \(\lambda = \frac{{4L}}{n} = \frac{{480}}{n}\), n = 1, 3, 5,…
OR
diagram showing \(\frac{3}{4}\) of a standing wavelength in the tube
\(1.6 = \frac{{4.80}}{n} \Rightarrow n = 3\)
OR
justification that \(\frac{3}{4} \times 1.6 = 1.2{\rm{m}}\)
Allow diagram showing \(\frac{3}{4}\) of a wavelength for MP1.
