NCEA Level 3 Physics 91523 (3.3) — page 1 of 6

SAMPLE ASSESSMENT SCHEDULE

Physics 91523 (3.3): Demonstrate understanding of wave systems

Assessment Criteria

Achievement / Achievement with Merit / Achievement with Excellence
Demonstrate understanding requires writing statements that typically show an awareness of how simple facets of phenomena, concepts or principles relate to a described situation. For mathematical solutions, relevant concepts will be transparent, methods will be straightforward. / Demonstrate in-depth understanding requires writing statements that will typically give reasons why phenomena, concepts or principles relate to given situations. For mathematical solutions the information may not be directly usable or immediately obvious. / Demonstrate comprehensive understanding requires writing statements that will typically give reasons why phenomena, concepts or principles relate to given situations. Statements will demonstrate understanding of connections between concepts.

Evidence Statement

NØ = No response; no relevant evidence.

One / Not Achieved / Achievement / Achievement with Merit / Achievement with Excellence
N1 / N2 / A3 / A4 / M5 / M6 / E7 / E8
Any ONE of the following. / Any TWO of the following. / Any TWO of the following. / Any THREE of the following. / Any TWO of the following. / Any THREE of the following. / Any THREE in (a) or any Two in (c). / Any SIX with at least TWO from each of (a) and (c).
(a) / The wave reflects off the fixed end.
The standing wave has seven nodes.
Interference occurs / Any TWO of the following:
·  The wave reflects off the fixed end and interferes with the in-coming wave.
·  Because the waves fit in the string./The string length is a multiple of λ
·  The standing wave has seven nodes including one node at each end – has six antinodes.
·  Amplitude of the wave grows
·  There are fixed nodes and antinodes
·  Nodes are caused by destructive interference and antinodes by constructive interference
·  When the wave reflects there is a phase change
·  A standing wave requires waves with the same frequency and amplitude to superimpose/interfere / Any TWO of the following:
·  The standing wave is the 6th harmonic.
·  There are points on the string at which the incident and reflected waves are always in antiphase; these points become nodes.
·  There are points on the string at which the incident and reflected waves are always in phase – these points become antinodes.
·  String length must be a multiple of λ/2
·  The string must be oscillated at a multiple of the fundamental frequency / Any THREE of the following:
·  The wave reflects off the fixed end and changes phase so that the fixed end is a node.
·  At λ/4 from the fixed point the incident and reflected waves are in phase.
·  λ/2 from the fixed end the incident and reflected waves are in antiphase, so this point becomes a node and there is no movement.
·  At the oscillator end, the string is also a node, due to similar reflections.
·  Because the length of the string is n(λ/2) long, the nodes and antinodes from both ends coincide, so the standing waves builds up.
(b) / ·  See Appendix One – sketch as shown or similar.
·  Wave velocity calculated to be 10ms-1 / Correct answer with some reasoning:
Eg, frequency of 6th harmonic (shown) is 5 Hz.
Frequency of 3rd harmonic is half of this, eg, 2.5 Hz.
(c) / Wavelength increases
Wave amplitude decreases / The standing wave disappears. / ONE of the following
·  The standing wave disappears because the wavelength is changed, so it does not fit.
·  The fundamental frequency and its harmonics are changed / Any TWO of the following
·  The ends remain nodes because the waves still reflect and there is a phase change.
·  The wavelength gets (1.05) longer, so the reflecting waves no longer perfectly cancel the incident waves along the string…
·  so there are no nodes and antinodes between the ends; these change with time.
·  The system does not resonate, so the amplitude does not build up in the same way and is much smaller.

NCEA Level 3 Physics 91523 (3.3) — page 1 of 6

Two / Not Achieved / Achievement / Achievement with Merit / Achievement with Excellence
N1 / N2 / A3 / A4 / M5 / M6 / E7 / E8
Any ONE of the following. / BOTH of the following. / Any ONE of the following. / BOTH of the following. / Any ONE of the following. / BOTH of the following. / (a) correct and any TWO points in (b). / BOTH of the following.
(a) / Uses /
A3 if power of ten error. (Eg, uses ) /
OR candidate uses:
/
BA = = 0.73 m
(b) / They will hear the sound getting louder and quieter, louder and quieter as they walk along the line.
Loud at the nodes and quiet at antinodes
Antinodes occur where a crest meets a crest and nodes occur where a crest meets a trough / Any ONE of:
·  Loud and quiet positions are due to constructive and destructive interference (between the sound waves from the speakers).
Constructive interference occurs when waves arrive at a point in phase
·  Destructive interference occurs when waves arrive out of phase / in antiphase.
The sound is loudest in the centre (A)
·  In a classroom reflected sounds interfere to makes the quiet spots less noticeable. / Any ONE points from:
·  time between a student hearing a maximum and a minimum intensity sound =

Eg, = 0.9 s.
·  The greatest contrast between the loud / quiet positions will be at the minima closest to A / There will be less contrast between loud / quiet at greater distances from A. / Candidate dicusses the cause of less contrast, relating it to amplitude / path difference / coherency / phase difference of sources
Eg, any ONE of:
·  The best cancellation will be close to the middle (either side of the central maximum) where the waves from each speaker are roughly equal in intensity. Waves which are in antiphase will cancel completely.
·  The cancellation will be less pronounced moving away from the middle because the waves will have differing amplitudes.
·  If the path difference is large there will be dissimilar amplitudes so cancellation won’t be clear.
·  Waves reflected from the walls of the classroom will have multiple path differences so there will be no clear constant phase difference to cause cancellation.
Three / Not Achieved / Achievement / Achievement with Merit / Achievement with Excellence
N1 / N2 / A3 / A4 / M5 / M6 / E7 / E8
Any ONE of the following. / Any TWO of the following. / Any ONE of the following. / BOTH of the following. / Any ONE of the following. / BOTH of the following. / Any ONE of the following. / BOTH of the following.
(a) / The ripples are caused by movement of the duck / Any TWO of:
·  the frequency of the duck is shown by how close the ripples are.
·  movement of the duck causes ripples to be off-centre
·  When the duck is stationary the ripples are circular, around one another
·  The ripples are circular because when the duck moves the wave fronts spread out at the same rate in all directions. / Any TWO of:
·  as the duck moves the ripples are centred in different positions.
·  If the wavelength is unchanged the duck isn’t moving.
·  movement of the duck horizontally cause the ripples to bunch up in front of the duck and spread out behind the duck / a short wavelength is produced ahead of the duck and a longer wavelength behind it. / Merit plus ONE of :
·  the frequency of the duck is shown by how close the ripples are -higher frequency causes closer ripples
·  a greater spacing between the ripples means they will be received less frequently by an observer
(b) / / Any ONE of:


Correct emission frequency: /

/





OR by working out that the observed wavelength = distance travelled in 1 s ¸ number of waves made in 1 s:


(speed of the wave relative to the duck - speed of the duck)
= 0.113 m s–1

Appendix One