NCEA Level 3 Physics (90520) 2010 — page 1 of 3
Assessment Schedule – 2010
Physics: Demonstrate understanding of wave systems (90520)
Evidence Statement
Q
Evidence
ONE
(a)(i) &
(ii)
(b)
(c)
f =
v
523
=
= 726 Hz
! 4 " 0.18
As it heats and becomes more humid v
increases, and for steam v again increases.
l remains the same, so λ is the same (for the
fundamental and all the overtones).
v
since f = , it increases with v.
!
If f increases, pitch increases.
Achievement
1
Correct answer –
allow schematic
diagrams and/or
labels.
2
Correct answer.
1
Pitch gets higher /
frequency
increases.
Achievement with
Merit
1
Explains that the
change in v causes:
Change in pitch due to
change in f.
OR
Change in pitch as λ is
constant.
OR
Change in f as λ is
constant.
Achievement
with Excellence
1
Complete
answer.
NCEA Level 3 Physics (90520) 2010 — page 2 of 3
(d)
fR = f
v
v + vt
2
343
343 + 15.2
! 343 + 15.2 $
f =#
' 540 = 564
343 &%
"
540 = f
fA = f
v
v ( vt
343
343 ( 15.2
! 343 ( 15.2 $
f =#
' 590 = 564
343 &%
"
590 = f
Uses average
value of frequency
(565 Hz) and
applies Doppler
equation with the
wrong sign
(+ to –).
OR
Uses correct form
of Doppler
equation but may
have incorrect
wave speed or be
unsure of how to
proceed.
2
Uses average value of
frequency (565 Hz)
and applies Doppler
equation correctly.
Eg v
f!= f
v ± vt
2
Calculates true
frequency twice
to show that
both
approaching and
receding have
same f.
OR correct
proof with
simultaneous
equations.
# f! &
vt = v %
" 1(
$ fR
'
# 565 &
= 343 %
"1
$ 590 ('
= 14.5 m s"1
OR:
v
f!= f
v ± vt
As the two equations give the same
frequency for a speed of 15.2 m s(1 ,
this must be the speed of the train.
OR
v
fR = f
v ( vt
#
f!&
vt = v % 1 "
f A ('
$
#
565 &
= 343 % 1 "
$ 540 ('
f R (v + vt )
v
v
fA = f
v + vt
f =
= 15.9 m s"1
OR correct method
with wave speed of
523, gives v = 23.1.
f A (v ( vt )
v
f R (v + vt ) f A (v ( vt )
=
v
v
540(343 + vt ) = 590(343 ( vt )
f =
vt = 15.17 m s –1 = 15.2 m s –1 (3 sf )
TWO
(a)
v
563
f = =
= 439.8 Hz
! 2 " 0.64
(b)(i)
They are exactly in antiphase and have the
same amplitude. (ii)
They are in phase but the amplitude of D is
much greater than C.
(c)
Shorter string produces higher frequency.
Frequency of the new string
f 2 = f1 + f beat = 440 + 20 = 460 Hz
v
!
563
f
l= =
=
= 0.612
2
2
2 " 460
#l = 0.640 $ 0.612 = 0.028 m
2
Correct method.
1
EITHER ONE
part completely
correct
OR BOTH correct
for amplitude (no
confusion with
displacement)
OR BOTH correct
for phase.
1
BOTH correct.
1
ALL correct.
2
Correct f2.
2
Correct length.
2
Correct answer.
NCEA Level 3 Physics (90520) 2010 — page 3 of 3
THREE If waves are in phase, the signal is strong /
(a)
intense because peaks reinforce peaks and
troughs reinforce troughs.
If they are out of phase then the signals cancel
(if they have the same amplitude and are
completely out of phase they will cancel
completely).
(b)
1.95 × 10–7 is exactly λ/4. A wave which
relects off a pit will travel an extra λ/2,
compared with a wave which reflects off the
surface.
Thus if the beam hits both the surface and a
pit, then the waves will be out of phase and
produce a weak signal.
Normally the reflections are in phase and
produce a strong signal.
As the disk rotates, it moves over pits and as
it travels over an edge, there will be a dip in
the intensity.
(c)
n! = d sin "
1
In phase – strong,
out of phase –
weak with no
reasons.
OR
Explains either
one but does not
decribe the other.
1
Correct answer with
reasoning, linking
phrase to constructive /
destructive
interference and
intensity.
1
Recognition that
1.95 × 10–7 is
exactly λ/4.
OR
Recognition that
the waves will
reflect in and out
of phase
producing strong
and weak signal as
the disk rotates.
1
Links the pit depth to
path difference for
destructive
interference and the
production of a weak
beam but does not
completely answer.
1
Complete
answer contrasts
signal at pit
edge to flat
surface.
2
Correct method (this is a
show question).
OR
Substitutes 15.18° to
show the wavelength is
4.36 m.
2
Correct θ.
OR
2
Correct answer.
$9
n! 436 # 10
=
= 0.2725
d
1.6 # 10$6
" = 15.8°
sin " =
(d)
2
tan !
366
=
tan15.8° 285
! = 19.97°
Uses !
= d sin "
with wrong θ.
Uses
x
L
twice to give
n! = d
" = d sin ! = 1.6 # 10$6 sin19.97°
" = 5.478 # 10$7 m
!gr = 5.60 " 10#7 m
Judgement Statement:
Achievement
5A, including at least two from each
criterion
Achievement
with Merit
4M, including at least one from each
criterion, + any 1A
Achievement
with Excellence
3E, including at least one from each
criterion, + any 1M
+ any 1A