NCEA Level 3 Physics 91523 (3.3) — page 1 of 5
SAMPLE ASSESSMENT SCHEDULE
Physics 91523 (3.3): Demonstrate understanding of wave systems
Assessment Criteria
Achievement
Achievement with Merit
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.
Achievement with Excellence
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.
Not Achieved
N1
One
Any ONE of the
following.
Achievement
N2
Any TWO of the
following.
The wave reflects off the fixed end.
The standing wave has seven nodes.
Interference occurs
(a)
A3
Any TWO of the
following.
Achievement with Merit
A4
Any THREE of
the following.
M5
Any TWO of the
following.
M6
Any THREE of
the following.
Any TWO of the following:
Any TWO of the following:
 The wave reflects off the fixed end
and interferes with the in-coming
wave.
 The standing wave is the
harmonic.
 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
Achievement with Excellence
E7
Any THREE in (a)
or any Two in (c).
E8
Any SIX with at
least TWO from
each of (a) and
(c).
Any THREE of the following:
6th
 There are points on the string at
which the incident and reflected
waves are always in antiphase;
these points become nodes.
 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.
 There are points on the string at
which the incident and reflected
waves are always in phase – these
points become antinodes.
 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.
 String length must be a multiple of
λ/2
 At the oscillator end, the string is
also a node, due to similar
NCEA Level 3 Physics 91523 (3.3) — page 2 of 5
 Nodes are caused by destructive
interference and antinodes by
constructive interference
 The string must be oscillated at a
multiple of the fundamental
frequency
 When the wave reflects there is a
phase change
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.
 A standing wave requires waves
with the same frequency and
amplitude to superimpose/interfere


(b)
Wavelength increases
Wave amplitude decreases
(c)
See Appendix One – sketch as
shown or similar.
Wave velocity calculated to be
10ms-1
The standing wave disappears.
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.
ONE of the following
Any TWO of the following
 The standing wave disappears
because the wavelength is
changed, so it does not fit.
 The ends remain nodes because
the waves still reflect and there is a
phase change.
 The fundamental frequency and its
harmonics are changed
 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 3 of 5
Not Achieved
N1
Two
N2
Any ONE of the
following.
Uses

Achievement
BOTH of the
following.
v
f
A3
Any ONE of the
following.
Achievement with Merit
A4
BOTH of the
following.
v
320

f 1300
 0.246 m

A3 if power of ten error. (Eg, uses
f  1.3 )
M5
Any ONE of the
following.
M6
BOTH of the
following.
d sin   n
(a) correct and
any TWO points
in (b).
E8
BOTH of the
following.
d sin   n
1  n 
1  0.5  0.246 
 n 
1  0.5  0.246 


sin

sin





sin


 d 
0.52 
 d 
0.52 
  13.7 (0.239 rad)
  13.7 (0.239 rad)
BA = 3.0  tan13.7 = 0.73 m
dx
L
n L
x
d
0.5  0.246  3.0

0.52
 0.71 m
n 
They will hear the sound getting
louder and quieter, louder and quieter
as they walk along the line.
Any ONE of:
Any ONE points from:

 time between a student hearing a
Loud at the nodes and quiet at
antinodes
(b)
E7
  sin 1 
OR candidate uses:
(a)
Achievement with Excellence
Antinodes occur where a crest meets
a crest and nodes occur where a crest
meets a trough
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)
maximum and a minimum intensity
sound =
distance calculated in (a)
time =
0.8
Eg, t 
d 0.71
= 0.9 s.

v
0.8
 The greatest contrast between the
loud /quiet positions will be at the
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.
NCEA Level 3 Physics 91523 (3.3) — page 4 of 5
 In a classroom reflected sounds
interfere to makes the quiet spots
less noticeable.
minima closest to A /There will be
less contrast between loud /quiet at
greater distances from A.
 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.
Not Achieved
Three
N1
Any ONE of the
following.
Achievement
N2
Any TWO of the
following.
The ripples are caused by movement
of the duck
(a)
A3
Any ONE of the
following.
Achievement with Merit
A4
BOTH of the
following.
M5
Any ONE of the
following.
M6
BOTH of the
following.
Achievement with Excellence
E7
Any ONE of the
following.
E8
BOTH of the
following.
Any TWO of:
Any TWO of:
Merit plus ONE of :
 the frequency of the duck is shown
by how close the ripples are.
 as the duck moves the ripples are
centred in different positions.
 movement of the duck causes ripples to be off-centre
 If the wavelength is unchanged the
duck isn’t moving.
 the frequency of the duck is shown
by how close the ripples are higher frequency causes closer
ripples
 When the duck is stationary the ripples are circular, around one another
 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.
 The ripples are circular because
when the duck moves the wave
fronts spread out at the same rate in
all directions.
 a greater spacing between the ripples means they will be received
less frequently by an observer
NCEA Level 3 Physics 91523 (3.3) — page 5 of 5
Any ONE of:
Correct emission frequency:
(b)
OR by working out that the observed wavelength = distance
travelled in 1 s  number of
waves made in 1 s:

0.4  v
 0.135
19 / 5
 0.135  19 
v
  0.4
5

(speed of the wave relative to the
duck - speed of the duck)
= 0.113 m s–1
Appendix One