| Date | November 2010 | Marks available | 3 | Reference code | 10N.3.SL.TZ0.A3 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Deduce | Question number | A3 | Adapted from | N/A |
Question
This question is about standing waves and organ pipes.
An organ pipe of length \(L\) is closed at one end. On the diagrams, draw a representation of the displacement of the air in the pipe when the frequency of the note emitted by the pipe is the
State one way in which a standing wave differs from a travelling wave.
(i) first harmonic frequency \({f_1}\).
(ii) second harmonic frequency \({f_2}\).
Use your answer to (b) to deduce an expression for the ratio \(\frac{{{f_1}}}{{{f_2}}}\).
State, in terms of the boundary conditions of the standing waves that can be formed in the pipe, the reason why the ratio of the higher frequencies of the harmonics to that of the first harmonic must always be an integer number.
Markscheme
no energy propagated in a standing wave;
the amplitude of a standing wave is not constant;
points along a standing wave are either in phase or out of phase with each other / OWTTE;
(i) antinode at open end node at closed end;
(ii) antinode at open end and node at closed end and one more node along pipe;
(judge by eye)
for \({\lambda _1} = 4L\) and for \({\lambda _2} = \frac{{4L}}{3}\);
\({f_1} = \frac{c}{{4L}}\) and \({f_2} = \frac{{3c}}{{4L}}\);
\(\frac{{{f_1}}}{{{f_2}}} = \frac{1}{3}\);
there must always be a node at the closed end and an antinode at the open end / there must always be an integer number of \(\frac{\lambda }{4}\);
Examiners report
Most candidates knew a difference between a standing and travelling wave.
Diagrams of the fundamental and second harmonic were often poor.
The manipulation of ratios defeated a lot of candidates.
Very few candidates recognised that there must always be either a node or antinode at each end of the pipes.
