9 Wave phenomena
9.1 Simple harmonic motion
Answers to 9.1 Displacement, velocity, and acceleration calculations 1
(Skills worksheet)
1)
a) Maximum displacement: 0.12 m
b) Maximum velocity: 3.8 ms-1
c) Maximum acceleration: 120 ms-2
d) Equation of motion: x = 0.12 sin(31t)
e) Velocity at t = 0.15 s: 0.24 ms-1
2) Period: 0.80 s
3) Length: 0.376 m
4) Velocity at 0.111 m: 24.7 ms-1
5)
6)
a) Equation of motion: x = 0.022 cos(280t)
b) Displacement at 0.014 s: -0.016 m
a) Maximum velocity: 0.277 ms-1
b) Maximum acceleration: 0.867 ms-2
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1
9 Wave phenomena
9.1 Simple harmonic motion
Answers to 9.1 Displacement, velocity, and acceleration calculations 1
(Skills worksheet)
1)
a) Maximum displacement: 0.12 m
b) Maximum velocity: 3.8 ms-1
c) Maximum acceleration: 120 ms-2
d) Equation of motion: x = 0.12 sin(31t)
e) Velocity at t = 0.15 s: 0.24 ms-1
2) Period: 0.80 s
3) Length: 0.376 m
4) Velocity at 0.111 m: 24.7 ms-1
5)
6)
a) Equation of motion: x = 0.022 cos(280t)
b) Displacement at 0.014 s: -0.016 m
a) Maximum velocity: 0.277 ms-1
b) Maximum acceleration: 0.867 ms-2
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1
9 Wave phenomena
9.1 Simple harmonic motion
Answers to 9.1 Energy transfer (Skills worksheet)
1)
a) Kinetic energy: 0.80 J
b) Speed: 0 ms-1
2) Potential energy / Kinetic energy: 1/15
3) Energy increase: 9 times
4)
a) Blue: velocity-time; Red: kinetic energy-time
b) Potential energy at 0.3 s: 0.50 (no units)
c) Potential energy graph has a period of 0.4 seconds, minimum 0.00, maximum 1.00, sin2x
shape. Total energy is just a straight line (equal to 1 at all times).
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1
9 Wave phenomena
9.2 Single-slit diffraction
Answers to 9.2 Single-slit diffraction 1 (Skills worksheet)
1) Wavelength: 536 nm
2) Angular width: 4.91o
3) Distance: 0.161 m
4) Slit width: 2.92 x 10-5 m
5) Slit width: 7.90 x 10-6 m
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1
9 Wave phenomena
9.2 Single-slit diffraction
Answers to 9.2 Single-slit diffraction 2 (Skills worksheet)
1)
a) Distance: 0.189 m
b) Width: 0.377 m
c) Angular width: 0.296 rad
d) Yes, to 0.283 m which is a factor of 1.5
2) Distance: 0.263 m
3) Difference in width: 0.0155 m
4) Maximum width: slightly less than 180o
5) Wavelength: 594 nm
6) Number of minima: 4
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1
9 Wave phenomena
9.2 Single-slit diffraction
Answers to 9.2 Single-slit diffraction 3 (Skills worksheet)
1) Distance: 0.594 m
2) Ratio: 1:1
3) Distance: 5.90 cm
4) One mark at the centre and then one mark at 2.95 cm and again at 5.90 cm on each side.
5) Slit width: 2.51 x 10-6 m
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1
9 Wave phenomena
9.3 Interference
Answers to 9.3 Diffraction grating (Skills worksheet)
1) First order angle: 22.3o
2) Second order angle: 49.5o
3) Maximum number of orders: 2
4) Lines per cm: 9288
5) Distance: 5.85 cm
6) Wavelength: 486 nm
7) Lines per cm: 5840
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1
9 Wave phenomena
9.3 Interference
Answers to 9.3 Thin film interference (Skills worksheet)
1) Minimum film thickness: 7.94 x 10-8 m
2) Wavelength: 620 nm
3) Thickness: 9.25 x 10-8 m
4) The second ray travels an extra 2 x 605 nm = 1210 nm. This 1210/484 = 2.5 cycles. This is
180o out of phase but so are the two emerging rays therefore they are in phase and
constructive interference takes place.
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1
9 Wave phenomena
9.4 Resolution
Answers to 9.4 Resolution skills worksheet 1
1) Angle of resolution: 2.82 x 10-7 rad
2) Minimum separation: 6.43 x 106 m
3) Maximum distance: 1.33 x 1010 m
4) Angular resolution: 2.52 x 10-7 rad
5) Diameter: 3.05 mm or 2.95 mm
6) Distances: 7300 m or 7060 m (7056 m)
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1
9 Wave phenomena
9.4 Resolution
Answers to 9.4 Resolution skills worksheet 2
1) Distance: 0.912 m
2) Yes; Distance: 1.27 m
3) λ1 / λ2 = r2 / r1
4) Distance: 488 m
5) Separation: 8.00 x 107 m
6) The distance between the two objects is greater than the diameter of the largest moon in the
solar system, therefore; the scientist has made a mistake. There are other reasons, such as
Jupiter is so bright compared with Ganymede that it would be impossible to distinguish
anything from such a great distance.
7) Lines per mm: 6033 (6000)
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1
9 Wave phenomena
9.5 Doppler effect
Answers to 9.5 Doppler effect skills worksheet 2
1) Frequency: 650 Hz
2) Frequency: 1560 Hz
3) Frequency: 598 Hz
4) Frequency: 388 Hz
5) Speed of sound: 359 ms-1
6) Speed of the car: 22.6 ms-1
7) Frequency: 533 Hz
8) The observer is driving away from the source
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1
9 Wave phenomena
9.5 Doppler effect
Answers to 9.5 Doppler effect skills worksheet 3
1) Frequency: 946 Hz
2) Frequency: 1740 Hz
3) Frequency: 988 Hz
4) Frequency: 439 Hz
5) Speed of sound: ? ms-1
6) Speed of car: 50.8 ms-1
7) Frequency: 594 Hz
8) The observer is not driving away from the source
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1
9 Wave phenomena
9.5 Doppler effect
Answers to 9.5 Doppler effect skills worksheet 4
1) Frequency: 459 Hz
2) Maximum wavelength: 656.008 nm
Minimum wavelength: 655.993 nm
3) Speed: 938 ms-1
4) Blood flow rate: 0.832 ms-1
5) Speed of galaxy: 6.7 x 106 ms-1 away
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1
E N D - O F -TO P I C Q U E S T I O N S
Solutions for Topic 9 – Wave phenomena (AHL)
1. a) acceleration is directly proportional to displacement; acceleration in opposite direction
to displacement
b) a = –ω2x
_
gradient of graph = –5 × 106 so ω = √5 × 106 =2236 rad s–1
2240 = 350 Hz
frequency = _
2π
c) amplitude = maximum displacement = 0.60 mm
2. a) (i) maxima or minima of curve (max acceleration at max or min displacement)
(ii) either point of intercept with time axis (maximum speed at zero displacement)
b) SHM part of circular path; centripetal force towards centre of circle = T – mg, therefore T > mg.
1 mv2
c) (i) potential energy mgh converted to kinetic energy _
2
_
so v = √ 2gh = 0.70 m s–1
mv2
(ii) T =
+ mg = 0.035 + 0.56 = 0.59 N
r
3. a) restoring force F = –kx (opposite direction to displacement)
0.07 EK /J
b) (i)
0.06
0.05
0.04
0.03
0.02
05
0.
04
03
0.
0.
02
0.
01
0
0.
04
0.
03
−
0.
02
−
0.
01
0.
−
−
−
0.
05
0.01
x/m
1 mv2 gives maximum velocity of 0.63 m s–1
(ii) EK= _
2
0.63
vmax = ω x0 where x0= 0.05 m so f = _
= 2.0 Hz
2π × 0.05
4. a) diffraction of light occurs when light passes through a narrow slit, causing waves to bend and
create an interference pattern
b)
I
C
d
2 × D × λ = 5.9 mm
c) width = _
a
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1
E N D - O F -TO P I C Q U E S T I O N S
5. a)
Q
Z
ϕW
b
Y
θ
X
P
screen
slit
b) path difference is half of a wavelength for destructive interference
b sinθ; use small angle approximation and rearrange to get answer
λ=_
c) ZW = _
2 2
2 × 450 × 10 = 6.0 × 10–3 rad
d) angular width __
0.15 × 10–3
–9
6. a) waves between A and B at same intensity with same spacing as original graph
450 × 10 gives θ = 21°
λ = __
b) sinθ = _
d 1.25 × 10–6
–9
d = ___
1
7. a) _
= 2.8 so maximum order in each direction is 2, plus zero order gives 5.
λ 6.0 × 105 × 590 × 10–9
b) second order peak will be wider and fainter than first order peak
8. a) (i) 180° or π
(ii) 0
620 × 10–9 × _
λ×_
1 =_
1 = 2.2 × 10–7 m
b) minimum thickness is _
2 n
2
1.4
intensity I
9. a) b)
B
A
distance
λ = 2.4 × 106
c) θ = 1.22 _
a
separation s = θ × d = 2.4 × 106 × 8.1 × 1016 = 2.0 × 1011 m
10. a) ratio of the wavelength of the light to the smallest difference in wavelength that can be
resolved by the grating
2000 = 10000 lines mm–1
b) (i) _
0.2
(ii) ∆λ = 0.2 nm
λ = 3280 which is greater than the resolvance, so no.
_
∆λ
11. wave speed remains the same, wavelength measured by observer is smaller as wave fronts are
closer together due to approaching sound source.
12. a) f’ is higher than f due to Doppler effect; observer is walking towards source so intercepts crests
of wavefront at higher rate than they are emitted
v
330
_
2
b) f ’ = f _
v – us 3.0 × 10 × 315 = 314 Hz
13. all lines shifted to the right to slightly higher wavelengths (redshift); shift is greater at higher
wavelengths
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2
IB Physics
Assessment paper answers: 9 Wave phenomena (AHL)
1. A
(1)
2. C
(1)
3. D
(1)
4. A
(1)
5. A
(1)
6. B
(1)
7. B
(1)
8. C
(1)
9. A
(1)
10. B
(1)
Total: 10 marks
1. a) Q is destructive interference since the difference in path lengths from the
speakers to Q must be a ½ λ difference;
(3)
Accept (n + 1/2) difference.
b) frequency = 2 Hz; time between loud points = 0.5 s;
wavelength of sound, λ = 330 / 360 = 0.917 m;
distance between loud point = λD / a = (0.917 x 10.0) / 4.0 = 2.29 m
speed = distance / time = 2.29 / 0.5 m s-1 = 4.58 m s-1 ≈ 5 m s-1
(6)
Accept solutions using nλ = d sin Ɵ.
The solution above is, of course, an estimation. If the answer attempts “full”
solution (ie does not resort to small angle estimations), but still gets
confused with the maths, full marks can be awarded if appropriate
progress is made or understanding shown.
Total: 9 marks
2. a) the phase difference between light leaving S1 and S2 is constant
(1)
Do not penalize the candidate if they state “has the same phase”.
b) to produce sufficient diffraction;
for the beams to overlap;
(2)
Total: 3 marks
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1
IB Physics
Assessment paper answers: 9 Wave phenomena (AHL)
3.
490 =
𝑓
𝑓
;
410
=
𝜈
𝜈
1 − 340
1 + 340
490
340 + 𝜈
=
410
340 − 𝜈
ν = 30 m s-1
OR
490 =
𝑓
𝜈
1 − 340
f = 450 Hz; justification of f = 450 Hz to get ν = 28 m s-1
OR
410 =
𝑓
1+
𝜈
340
f = 450 Hz; justification of f = 450 Hz to get ν = 33 m s-1
(4)
Total: 4 marks
4. a) the maximum of one diffraction pattern is coincident with the first minimum
of the other;
OR
(2)
b) Ɵmin = 1.22(λ/D);
with small angle approximation: Ɵ = (s/150 m);
equate Ɵ = 1.22(λ/D) = 1.22(590 x 10-9/5.0 x 10-3) to get s = 2.2 cm;
(3)
Award (2) max if 1.22 factor is omitted.
Total: 5 marks
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2
IB Physics
Assessment paper answers: 9 Wave phenomena (AHL)
5. a) each element of the slit acts as a point source of light;
the light from these sources interfere;
there will be a zero of intensity (on the screen) when the sum of the path
differences between the sources is an integral number of half wavelengths /
a maximum when an integral number of wavelengths;
(3)
b)
central maximum same intensity as single slit maximum;
two other maximum either side about half-intensity of central maximum;
Award (1) max if lines do not touch the x-axis.
(2)
There is no need to show maxima within secondary maxima. Do not
penalize responses if more than two maxima are shown but they must be
symmetrical and with realistic relative intensities.
Total: 5 marks
6. a) changes by π / 180°;
(2)
b) use of 2nd cosϕ = m +1/2 λ
n = 1 cosϕ = 1 and m changes by 1, therefore d = λ/2 = 280 nm
(2)
Total: 4 marks
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3
Answers to exam-style questions
Topic 9
Where appropriate, 1 ✓ = 1 mark
1 B
2 D
3 A
4 D
5 A
6 B
7 A
8 C
9 A (The question should have referred to the wavelength in air)
10 C
11 a In simple harmonic motion the acceleration is opposite to and proportional to the displacement from the
equilibrium position. ✓
This means that a graph of acceleration against time should a straight line through the origin with a negative slope. ✓
Which is what this graph is. ✓
b i The amplitude is 2.6 cm. ✓
12
−1
ii The gradient is −ω 2 = −
−2 ⇒ ω = 15.19 rad s ✓
5.2 × 10
15.19
ω = 2π f ⇒ f =
= 2.4 Hz ✓
2π
1
1
c i E max = mω 2 x 02 = × 0.25 × 15.19 2 × (2.6 × 10 −2 )2 ✓
2
2
−2
E max = 1.9479 × 10 ≈ 1.9 × 10 −2 J ✓
1
ii E K = E P ⇒ E K = E max ✓
2
1
E K = × 1.9497 × 10 −2 = 9.75 × 10 −3 J ✓
2
1
× 0.25 × v 2 = 9.75 × 10 −3 ⇒ v =
2
2 × 9.75 × 10 −3
= 0.279 ≈ 0.30 m s −1 ✓
0.25
d Correct shape of parabola. ✓
Correct intercepts. ✓
E × 10–2 J
2.0
1.5
1.0
0.5
–3
–2
–1
0
1
2
3
x / cm
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 9
1
12 a i Light diffracting from each slit arrives at the screen. ✓
At those positions where the phase difference between the 2 waves is 0 the resulting amplitude is twice that
of the wave from just one slit and we have bright fringes (constructive interference). ✓
ii The separation of the bright fringes is given by s =
sd
λD
and so λ = . ✓
d
D
1.86 × 10 −2 × 0.120 × 10 −3
✓
3.60
λ = 6.20 × 10 −7 m ✓
b Correct overall shape. ✓
Correct peak intensity. ✓
Correct separation of fringes. ✓
λ=
I / Wm–2
80
60
40
20
–20
–10
0
10
20
θ / mrad
nλ
2 × 6.2 × 10 −7
= 1.462 × 10 −6 m = 1.462 × 10 −3 mm ✓
=
sin θ
sin 58°
1
Hence number of rulings per mm is
=684 ✓
1.462 × 10 −3
c i d sin θ = nλ ⇒ d =
ii We must have that 1.462 × 10 −6 × sin 58° = nλ so that nλ = 1.2398 × 10 −6 m. ✓
n = 1 does not lead to a visible wavelength. ✓
1.2398 × 10 −6
We cannot have n = 2 so we try n = 3 to find λ =
= 4.13 × 10 −7 m which fits the visible
3
spectrum. ✓
No other value of n gives a visible wavelength. ✓
13 a Parallel reflected rays in red. ✓
Correct refraction of one of the rays. ✓
incident
MgF2
d
glass
b At reflection point between air and magnesium fluoride. ✓
At reflection point between magnesium fluoride and glass. ✓
c At normal incidence the path difference is 2d and the phase difference due to reflection is zero. ✓
1
Hence for destructive interference 2dn = (m + )λ . ✓
2
λ 5.0 × 10 −7
Giving for the least thickness (m = 0) d =
=
= 9.1 × 10 −8 m. ✓
4n
4 × 1.38
2
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 9
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
14 a The number of secondary maximum is 2 less than the number of slits. ✓
And we have 2 secondary maxima. ✓
b i The secondary maxima becomes less pronounced. ✓
The primary maxima become brighter. ✓
The primary maxima become narrower. ✓
ii The separation between the primary maxima increases. ✓
656.45 + 656.27
c The average wavelength is
= 656.36 nm. ✓
2
656.36
λ
From
= 2 × N. ✓
= mN we have that
656.45 − 656.27
∆λ
N = 1823 ✓
15 a The first minimum is at 0.175 mrad. ✓
λ
5.0 × 10 −7
λ
And so from θ = 1.22 we find b = 1.22 = 1.22 ×
= 3.49 × 10 −3 m. ✓
b
θ
0.175 × 10 −3
b i Same shape. ✓
With maximum coinciding with first minimum of the other pattern. ✓
Intensity
1.0
0.8
0.6
0.4
0.2
–0.2
0.0
0.2
0.4
0.6
θ /mrad
−2
ii The angular separation of the two sources is 3.0 × 10 where D is their distance from the slit. ✓
D
3.0 × 10 −2
3.0 × 10 −2
According to Rayleigh,
= 171 ≈ 170 m. ✓
= 0.175 × 10 −3 giving D =
D
0.175 × 10 −3
16 a The change in observed frequency when there is relative motion between the source and the observer. ✓
b Circular wavefronts. ✓
Bunching in front of the source. ✓
stationary
observer
source moving
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 9
3
c Ultrasound is directed at moving particles in the blood stream and the reflection is recorded. ✓
From the frequency shift it is possible to measure the speed of blood flow. ✓
d The speed of the point on the disc is 2π × 0.20 = 10.0 m s −1. ✓
1
8
340
The frequencies received range from
× 2400 Hz = 2331 ≈ 2300 Hz when source moves away from
340 + 10
observer, ✓
340
× 2400 Hz = 2473 ≈ 2500 Hz when source moves towards the observer. ✓
340 − 10
The wavelengths correspondingly vary from 340 = 0.137 ≈ 0.14 m to 340 = 0.146 ≈ 0.15 m. ✓
2473
2331
to
4
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 9
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
Answers
to Questions
The speakers
in a stereo system vibrate, but usually in a very complicated way since many notes are
being sounded at the same time.
The
blades
in an
electricwhen
shaver
vibrate,
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piano
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very complicated way since many notes are
The pistons
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being
sounded
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the
same
time.
The free end of a diving board oscillates after a diver jumps, in approximately SHM.
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The acceleration
pistons in a car
inoscillator
approximately
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of aengine
simpleoscillate,
harmonic
is zeroSHM.
whenever the oscillating object is at the
Conceptual
Questions
The free endposition.
of a diving board oscillates after a diver jumps, in approximately SHM.
equilibrium
1.
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11:
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CHAPTER
11: ofVibrations
and
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its motion – the
top Waves
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Answers
to examples
Questions
1. Give
of everyday
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Which
exhibit
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atofleast
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The some
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can
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ason
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–
and
simple
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motion
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at
the
extremes
1. The blades
in an electric shaver vibrate, approximately in SHM.
Answers
to Questions
at
the
extremes
of
its
motion
–
the
top
and
bottom
of
the
stroke
–
which
is
the
same
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The speakers in a stereo system vibrate, but usually in a very complicated way since many notes are
harmonic
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exerted
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is similar to Fig. 11-6.
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the
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mass means
a smaller
frequency.About
Thus1/3
the the
truemass
frequency
2. The
acceleration
simple
oscillator is zero whenever the oscillating object is at the
2a the
m
spring
contributesofto
totalharmonic
mass value.
3. equilibrium
Thesmaller
motionthan
of the
can be
approximated
as simple
harmonic.
all, the piston
will have
position.
be
thepiston
“massless
spring”
approximation.
And
since theFirst
true of
frequency
is smaller,
the a
constant
period
while
the
engine
is
running
at
a
constant
speed.
The
speed
of
the
piston
will
be
true period will be larger than the “massless spring” approximation. About 1/3 the mass of the zero
vmax
A bottom
k mas. simple
5.
The
maximum
speed
is given
by
Various
combinations
changing
A,ink,simple
and/or
m a
at themotion
extremes
of piston
its
motion
the
top
and
of the harmonic.
stroke
– which
isofthe
as
3. The
of the
can –be
approximated
Firstof
all,same
the piston
will have
spring
contributes
to the
total
mass
value.
3. How
could
you
double
the
maximum
speed
of
a
simple
harmonic
oscillator?
harmonic
motion.
There
is
a
large
force
exerted
on
the
piston
at
one
extreme
of
its
motion
–
the
can resultperiod
in a doubling
the maximum
speed.
For example,
k and
m areofkept
constant,
constant
while theofengine
is running
at a constant
speed.if The
speed
the piston
willthen
be zero
combustion
the
fuel
mixture
– and
harmonic
the klargest
force
at
extremes
of
doubling
theofamplitude
will double
thesimple
maximum
speed.
Or, ifhas
and
are
kept
constant,
then
at
the extremes
of its
motion
– the
top
and
bottom
of themotion
stroke
–Awhich
is the
same
as the
in simple
A kin m
5. The
maximum
speed the
is given
by vmaxmoves
. Various
combinations
of changing
A,dimension
k, and/or m
the motion.
Also,
crankshaft
a circle,
component
of motion
reducing
the
mass as
to
one-fourth
its force
original
value
will
double
the
maximum
speed.
Note
that
harmonic
motion.
There
is a large
exerted
on
theits
piston
at one
extreme
ofinitsone
motion
– the is
can
result into
athe
doubling
ofItthe
maximum
speed.
For example,
if k the
andlargest
m are kept
then
transferred
is similar
to Fig.
11-6.
combustion
of
thepiston.
fuel mixture
– and simple
harmonic
motion has
forceconstant,
1at the
k extremes of
doubling
the
amplitude
will
double
the
maximum
speed.
Or,
if
A
and
k
are
kept
constant,
then
f
changing
either
k
or
m
will
also
change
the
frequency
of
the
oscillator,
since
.
the motion. Also, as the crankshaft moves in a circle, its component of motion in one dimension is
m that
reducing
to has
one-fourth
its mass
original
will double
the than
maximum
speed.
Note
4. transferred
Since the the
real
spring
mass,
the
thatvalue
is moving
is greater
the mass
at 2the
end
of the
tomass
the
piston.
It is similar
to Fig.
11-6.
1 k
1 k
f reading
changing
either
m
will
change
the
frequency
ofisthe
oscillator,
since
.of the
6. Since
The
scale
reading
will
oscillate
with
damped
about
anthan
equilibrium
of 5.0
kg,will
f k or
spring.
,also
a larger
mass
means
a smaller
frequency.
Thus
the
4.
theSince
real
spring
has
mass,
the
mass
that
isoscillations
moving
greater
the
mass
attrue
thefrequency
end
2
mof 5.0 kg (so the range of readings is initially from 0.0 kg2and m
with an initial amplitude
10.0 kg).
1 spring
k and
be smaller
thanfin
thethe
“massless
spring”
approximation.
since the
true
frequency
is
smaller,
Due
to friction
scale
mechanism,
the And
oscillation
amplitude
willtrue
decrease
over the
time,
spring.
Since
, a larger
mass
means a smaller
frequency.
Thus
the
frequency
will
6.Chapter
The
scale
reading
will
oscillate
with
damped
oscillations
about
an
equilibrium
reading
of
5.0
kg,
true
period
will
be
larger
than
the
“massless
spring”
approximation.
About
1/3
the
mass
of
the
2
m
eventually
coming to rest at the 5.0 kg mark.
11
Vibrations and Waves
4. Ifwith
a pendulum
clocktoisthe
accurate
at
level,
will itofgain
or lose
when
taken
higher
Why?
ancontributes
initial
of 5.0spring”
kgsea
(so
the range
readings
is time
initially
from
0.0 to
kgaand
10.0altitude?
kg).
spring
total
mass
value.
be
smaller
thanamplitude
the “massless
approximation.
And since
the true
frequency
is smaller,
the
Due period
to friction
the
spring
and
scale
mechanism,
the
oscillation
amplitude
will
decrease
time,
7. true
The
of ain
pendulum
clock
inversely
proportional
to the square
root 1/3
of
g,the
bymass
Equation
will
be
larger
than
theis
“massless
spring”
approximation.
About
ofover
the11-11a,
eventually
coming
to
rest
at
the
5.0
kg
mark.
thetoenergy
istobeing
received,
and
energy
still decrease
received of
in changing
spite
of the
it m
is a
contributes
the
total
mass
value.
vmax
A k the
m .value
5. spring
The
istaken
given
by
Various
A, barrier,
k, and/or
Tlocation
2maximum
Lwhere
g . speed
When
high
altitude,
of giscombinations
will
(by
a small
amount),
which
good indication that the energy is being carried by waves. If the placement of the barrier stops the
can result
in
of clock
the maximum
speed.
Forlong,
example,
ifsquare
k and
m
areofkept
constant,
then
the
period
increase.
If
period
is too
the
clock
isout
running
slow
and
will
losebe
7. means
The
period
ofa adoubling
pendulum
is the
inversely
proportional
to the
root
g, by
Equation
11-11a,
energy
transfer,
itwill
could
be that
energy
is being
carried
by
particles.
It so
could
also
vmaxthe
A ktransfer
m . Various
5. The
maximum
speed
is given
by the
combinations
of changing
A,
k,
and/or
doubling
the amplitude
will double
maximum
speed. Or,
if A and k are
kept constant,
then m
time.
transfer
is being
carried
out with
waves
whose
wavelength
isamuch
smaller
than
the
Tthatresult
2the energy
Lthe
. doubling
When
taken
to maximum
high
altitude,
the value
of
g willthe
small
amount),
which
reducing
mass
to one-fourth
its original
value
willexample,
double
Note
that
can
ingaof
of the
speed.
For
ifdecrease
kmaximum
and m (by
are speed.
kept
constant,
then
dimensions
the barrier.
means thethe
period
will increase.
If the
is toospeed.
long, the
is running
slow
will lose
doubling
amplitude
will double
theperiod
maximum
Or,clock
if A and
k are kept
constant,
1and so
k then
f
changing
either
k
or
m
will
also
change
the
frequency
of
the
oscillator,
since
.
time.
Calculation-based
Questions
reducing the mass to
one-fourth its original value will double the maximum speed. Note that
2 laws
m as they
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright
1 fromk the
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing
Solutions
to
Problems
f
k or m SHM
will also
the frequency
of the
oscillator,
since
. in one period?
publisher.
1. Ifchanging
a particleeither
undergoes
withchange
an amplitude
of 0.18m
what
is the total
distance
it travels
2
m
6. The scale reading will oscillate with damped oscillations
about an equilibrium reading
of 5.0 kg,
269
©1.2005with
Pearson
Education,
Inc., Upper
Saddle
River,
All
reserved.
This material
is protected
under
an
initialwould
amplitude
offour
5.0
kg NJ.
(sothe
therights
range
of readings
initially
0.0
kgAand
10.0
0 to all
x copyright
x as they
0kg).
x is A
The
particle
travel
times
amplitude:
from
to x from
tolaws
to
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
Due
to
friction
in
the
spring
and
scale
mechanism,
the
oscillation
amplitude
will
decrease
over
time,
6.
The
scale
reading
will
oscillate
with
damped
oscillations
about
an
equilibrium
reading
of
5.0
kg,
publisher.
x A . So the total distance 4 A 4 0.18 m
0.72 m .
eventually
coming
to restofat5.0
thekg
5.0(so
kgthe
mark.
with
an initial
amplitude
range269
of readings is initially from 0.0 kg and 10.0 kg).
Due to friction in the spring and scale mechanism, the oscillation amplitude will decrease over time,
2. eventually
Theperiod
springcoming
constant
theatclock
ratio
of
applied
force
to displacement.
7.
of
a pendulum
is
proportional
to the square
of g, by Equation
11-11a,
to isrest
the 5.0
kg
mark.
2. IfThe
a pendulum
undergoes
exactly
20inversely
oscillations
in 34.6s, calculate
(a) itsroot
frequency;
(b) its angular
frequency.
F
180
N
75
N
105
N
2
# L g . When
#
T
2
taken
to
high
altitude,
the
value
of
g
will
decrease
(by
a
small
amount),
which
5.3 10
N m
=k $ = of
= 0.578𝐻𝑧;
(b) 𝜔 = 2𝜋𝑓
= 2𝜋(0.578)
3.63𝑟𝑎𝑑/𝑠.
7. (a)
The𝑓 period
a0.85
pendulum
clock
inversely
proportional
to the=square
root of g, by Equation 11-11a,
x (&'.)/+,.,)
m 0.65
m is0.20
m
means the period will increase. If the period is too long, the clock is running slow and so will lose
T 2 L g . When taken to high altitude, the value of g will decrease (by a small amount), which
time.
3. AThe
constant
is found
from
the ratio
applied oscillations
force to displacement.
3.
childspring
on
swing
went
through
five of
complete
10.4s. (a)slow
What
was
means
theaperiod
will
increase.
Ifexactly
the 2period
is too long,
the clock in
is running
and
so the
willperiod?
lose (b)
68
kg
9.8
m
s
What
F frequency?
mg
time. was the
k #,.'
1.333 105 N m
# 3 #
x
x
5
10
m
(a)Pearson
𝑇 = Education,
= 2.08𝑠;
(b)Saddle
𝑓 = River,
= NJ. All
= 0.481𝐻𝑧
© 2005
Inc., Upper
rights reserved. This material is protected under all copyright laws as they
>.,,
$
+.,?
currently
exist.frequency
No portion ofof
thisoscillation
material may is
be found
reproduced,
in any
or by
any means,
without
permission
in writing from the
The
from
theform
total
mass
and the
spring
constant.
publisher.
5 All rights reserved. This material is protected under all copyright laws as they
© 2005 Pearson Education,
River,10
NJ.
1 Inc.,
k Upper
1 Saddle
1.333
N m
269 by any means, without permission in writing from the
currently exist. No
form orHz
f portion of this material may be reproduced, in any1.467
1.5 Hz
publisher.
2
m 2
1568 kg
269
4.
(a) The spring constant is found from the ratio of applied force to displacement.
4. A mass oscillates between two springs with a frequency of 1.4Hz. (a) What is the angular frequency? (b)
What is its acceleration when its displacement is 1.0cm, 4.0cm, and when it passes through the equilibrium
position?
(a) 𝜔 = 2𝜋𝑓 = 2𝜋(1.4) = 8.8𝑟𝑎𝑑/𝑠 ; (b) 𝑎 = −𝜔+ 𝑥 = −(8.8)+ (0.010) = −0.77𝑚/𝑠 + ; 𝑎 = −𝜔+ 𝑥 =
−(8.8)+ (0.040) = −3.1𝑚/𝑠 + ; 𝑎 = −𝜔+ 𝑥 = −(8.8)+ (0.0) = 0𝑚/𝑠 + .
5. An oscillating mass is set in motion with SHM. It is at its maximum displacement of 12cm when a stopwatch
is started, and its period of oscillation is 2.4s. Calculate (a) the displacement after 3.3s; (b) its maximum
speed; (c) its speed after 5.6s; (d) its speed when its displacement is 8.8cm.
(a) 𝑥 = 𝑥, cos 𝜔𝑡 and 𝜔 = 2𝜋⁄𝑇, with 𝑥, = 0.12𝑚, 𝑇 = 2.4𝑠 and 𝑡 = 3.3𝑠.
+O
𝑥 = 𝑥, 𝑐𝑜𝑠 N+.'' × 3.3Q = 0.12 × −7.04 = −0.084𝑚
+O
(b) 𝑣, = 𝜔𝑥, = N+.'Q × 0.12 = 0.31𝑚/𝑠
(c) 𝑣 = −𝑣, sin 𝜔𝑡 with 𝑣, = 0.31𝑚/𝑠, 𝑡 = 5.6𝑠
2𝜋
𝑣 = 0.31 × sin U
× 5.6V = 0.27𝑚/𝑠
2.4
(d) 𝑣 = ±𝜔X(𝑥,+ − 𝑥 + ), with x = 0.088m
2𝜋
𝑣 = U V × X(0.12+ − 0.088+ ) = 0.21𝑚/𝑠
2.4
Topic 4.1b Kinematics of SHM (II) Problems
Calculation Based
1.
A mass attached to a spring is stretched a distance A from the equilibrium position and released. State the distance
from the equilibrium at which the acceleration becomes half of its initial maximum acceleration. [1 mark]
2.
A 0.60 kg mass vibrates on a horizontal spring according to the equation: 𝑥 = 0.45 cos 6.40𝑡, where x is in metres and t
is in seconds. Determine:
a. The amplitude. [1 mark]
b. The frequency. [1 mark]
c. The Period. [1 mark]
d. The spring constant, k [2 marks]
3.
The displacement of a particle executing SHM is given by y = 5.0cos (2t) where y is in millimetres and t is in seconds.
Calculate:
a. the initial displacement of the particle,
b. the displacement at t = 1.2s,
c. the time at which the displacement first becomes -2.0mm,
d. the displacement when the velocity of the particle is 6.0mm/s.
4.
An object oscillates according to the equation: 𝑣 = −0.12 cos 2.0𝑡, where v is in m/s and t is in seconds. Determine the
expression for the displacement as a function of time. [2 marks]
By inspection: 𝜔 = 2.0𝑠 23 .
And, 𝑣456 = ±𝜔𝐴 = −0.12 ⟹ ±𝐴 =
:.3;
;.:
= 0.06𝑚.
Therefore, 𝑥 = −0.06 sin(2.0𝑡) as velocity function is a negative cosine function.
A particle undergoes SHM. Its position against time is displayed using the following graph:
3.0
Displacement (m)
5.
π
-3.0
Time (s)
Using the information provided on the graph:
a. State the amplitude. [1 mark]
b. State the Period. [1 mark]
c. Calculate the angular frequency. [1 mark]
d. State the expression for the displacement as a function of time. [1 mark]
(a) From the graph: 3m
(b) The time taken for one oscillation is π seconds.
(c) 𝑇 =
;C
D
⟹𝜔=
;C
E
= 2𝑠 23
(d) 𝑓(𝑥) = 3.0 cos(2𝑡)
6.
A particle undergoes SHM with amplitude 4.0mm and angular frequency of 2.0 s-1. At t = 0, the displacement is
H.:
√;
mm.
Determine the expression for the displacement in the format:
𝑥 = 𝐴 cos(𝜔𝑡 + 𝛿).
[3 marks]
(Hint: to determine the phase difference, set t = 0).
Topic 9.2 and 9.4 Diffraction and Resolution Problems
Conceptual Questions
(These questions are not in an IB style but instead designed to check your understanding of the concept of this topic. You
should try your best to appropriately communicate your answer using prose)
1. For diffraction by a single slit, what is the effect of (a) increasing the slit width, (b) increasing the
wavelength?
2. By what factor could you improve resolution, other things being equal, if you use blue light (wavelength =
450nm) rather than red (wavelength = 700nm)?
3. Give at least two advantages for the use of large reflecting mirrors in astronomical telescopes.
4. Atoms have diameter about 10-8cm. Can visible light be used to “see” an atom? Explain.
Calculation-based Questions
5. If 580nm light falls on a slit 0.0440mm wide, what is the full angular width of the central diffraction peak?
6. When blue light of wavelength 440nm falls on a single slit, the first dark bands on either side of the centre
are separated by 55.0°. Determine the width of the slit.
7. A single slit of width 1.50μm is illuminated with light of wavelength of 500.0nm. Find the angular width of
the central maximum.
8. Microwaves of wavelength 2.80cm fall on a slit and the central maximum at a distance of 1.0m from the slit
is found to have a half width (i.e. distance from the middle central maximum to the first minimum) of 0.67m.
Find the width of the slit.
9. From the information shown determine the wavelength used to obtain the single-slit diffraction pattern
shown. The screen is 0.60m from the slit and the slit width is 2.30cm. What kind of wave is most likely used?
2
=
;
( d1 - d 2 )
( 2.61cm )
r
which
r = 0.19cm.
=resolution, limit
10. What is the
angular
(ingives
degrees)
set by diffraction for a 254cm diameter mirror of the Mt.
(14.8cm ) (1.1cm )
d2
Wilson telescope? (wavelength = 550nm).
48. The minimum angular resolution is
(1.22 ) ( 550 ì 10- 9 m )
1.22l
q=
=
= 2.64 ì 10- 7 rad = (1.51 ì 10- 5 ) è.
D
(100in )( 0.0254 m in )
49. Two
The stars
angular
the eye,
whichresolved
is the required
resolution
usingHow
the telescope,
is the stars? Assume
11.
15 resolution
light yearsof
away
are barely
by a 55cm
telescope.
far apart are
-3
0.10 ì 10
d
) =the
the wavelength
=(550nm
and m
that
resolution is limited by diffraction.
4.0 ì 10- 4 rad.
qeye = eye =
-2
Leye
( 25 ì 10 m )
The resolution without the telescope is
12. How far away
can( 7.0
a human
two car headlights 2.0m apart? Consider only diffraction effects
km ) eye distinguish
d
q= =
= 1.82 ì 10- 5 rad.
5
and assume
eyeì pupil
diameter
is 5.0mm and a wavelength to 550nm.
L the
10 km
( 3.84
)
If we ignore the inverted image, the magnification is
qeye f o
6
10
M = and =Moon
;
13. The Earth
q
f e are separated by about 400x10 m. When Mars is 8x10 m from Earth, could a person
standing on Mars
resolve the Earth and its Moon as two separate objects without a telescope? Assume pupil
4.0 ì 10- 4 rad ) ( 2.0 m )
(
diameter of 5.0mm and
550nm.
= a wavelength
, whichofgives
f e = 0.091m = 9.1cm.
(1.82 ì 10- 5 rad ) fe
The resolution limit is
(1.22 ) ( 5.5 ì 10- 7 m )
1.22l
q=
=
= 6.1 ì 10- 6 rad.
D
( 0.11m )
50. The minimum angular resolution is
1.22l
.
q=
D
The distance between lines is the resolving power:
1.22l f
RP = f q =
= 1.22l ( f -stop ) .
D
f
we have
For
2
RP2 = (1.22 ) ( 550 ì 10- 9 m ) ( 2 ) = 1.34 ì 10- 6 m = 1.34 ì 10- 3 mm, so the resolution is
1
1
=
= 746lines mm.
RP2 (1.22 ì 10- 3 mm )
For
f
we have
16
RP2 = (1.22 ) ( 550 ì 10- 9 m ) (16.7 ) = 1.12 ì 10- 5 m = 1.12 ì 10- 2 mm, so the resolution is
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
246
1.00))sin
sin45
45è=
è=((1.619
1.619))sin
sinqqbb22,, which
which gives
gives qqbb22 ==25.90
25.90èè..
((1.00
16. For
the
refraction
at
the
first
surface,
we
have
We find
find the
the angle
angle of
of incidence
incidence at
at the
the second
second surface
surface from
from
We
n(90
sin
q
=
n
sin
q
;
air
a
b
90
è
q
+
90
è
q
+
A
=
180
è
,
which
gives
( è - qbb)) + ((90è - qcc)) + A = 180è, which gives
sin- 45
1.642è)-sin25.51
qb1 , è=
which
gives
(qq1.00
=)A
q è=
= (60.00
34.49
è; qb1 = 25.51è;
cc11 = A - qbb11 = 60.00è - 25.51è= 34.49è;
1.619è) sin
qb 2 , è=
which
gives
(q1.00
=) sin
A - 45
q è==(60.00
- 25.90
è=34.10
34.10
q
èè.. qb 2 = 25.90è.
cc22 = A - qbb22 = 60.00è - 25.90
We
the angle at
of
incidence
the second
surface from
Forfind
the refraction
refraction
at the
the
secondatsurface,
surface,
we have
have
For
the
second
we
Calculation-based
Questions
sin
èq-qccq==b )nnair
+
90qqèdd-;; qc ) + A = 180è, which gives
sin
(nn90
(sin
air sin
Topic 9.3 Diffraction Problems
q((1.642
qb34.49
= 60.00
- 25.51
è; gives
sin
34.49
è=
1.00
sinqqè=
,34.49
which
giveswhat
68.4
from
the normal;
normal;
è=
qqdd11 ==is68.4
from
the
))-sin
c1.642
1 = Alight
1falls
dd11, which
1. If 580-nm
on((èa1.00
slit))sin
0.0440mm
wide,
theèèfull
angular
width of the central diffraction
qc1.619
qb34.10
è
25.90
è=
34.10
è
.
2 = A -sin
2 = 60.00
peak?
è=((1.00
1.00))sin
sinqqdd22,, which
which gives
gives qqdd22 ==65.2
65.2èèfrom
from the
the normal.
normal.
((1.619 ))sin 34.10è=
For the refraction at the second surface, we have
[2 Marks]
n sin qc = nair sin q d ;
17. We
We find
find the
the angle
angle to
to the
the first
first minimum
minimum from
from
17.
è= (1.00 ) sin q--99d 1 , which gives q d 1 = 68.4èfrom the normal;
(1.642 ) sin 34.49
580 ìì 10
10 m
m))
mll ((11))((580
m
sinqq1min
=
0.0132, so
soqq1min ==0.755
0.755èè..
sin
==
==0.0132,
1min =
, ))which gives q1min
(1.619 ) sin 34.10
(1.00 )ììsin
D è=
2m
d 2 = 65.2èfrom the normal.
D
0.0440
10q--3d3m
10
((0.0440
Thus the
the angular
angular width
width of
of the
the central
central diffraction
diffraction peak
peak is
is
Thus
17. We find the angle to the first minimum from
D
q
=
2
q
=
2
0.755
è
=
1.51
è
.
Dq11 = 2q1min
è.
1min = 2 ((0.755è)) = 1.51
(1) ( 580 ì 10- 9 m )
ml
sin q1min =
=
= 0.0132, so q1min = 0.755è.
-3
Dthe
0.0440
ì
10
m
18.
The
angle
from
the
central
maximum
to
theisfirst
first
minimum is
is wide.
17.5èè..If the angle between the first dark
18.
The
angle
from
central
maximum
to
the
minimum
17.5
(
)
2. Monochromatic light falls on a slit that
2.60x10-3mm
We
find
the
wavelength
from
We
find
the
wavelength
from
Thus theon
angular
of the
peak
is
fringes
eitherwidth
side of
thecentral
centraldiffraction
maxim um
is 35.0°
(dark fringe to dark fringe), what is the
Dsin
sinqq1min
=m
mll ;;
D
1min =
Dq1 = 2q1min
= 2 ( 0.755
wavelength
of--66light
used?è) = 1.51è.
2.60 ìì 10
10 m
m))sin
sin((17.5
17.5èè)) ==((11))ll ,, which
which gives
gives ll ==7.82
7.82 ìì 10
10--77m
m == 782
782nm.
nm.
((2.60
[2 Marks]
18. The angle from the central maximum to the first minimum is 17.5è.
19.
For
constructive
interference
19. We
Forfind
constructive
interference
from the
the single
single slit,
slit, the
the path
path difference
difference is
is
the wavelength
from from
DDsin
==((=
m
sinqq1min
mm++l 12;12))ll ,, m
m ==1,1, 2,
2, 3,
3, ...
.....
-6
For
first
fringe
from
the
maximum,
have
For the
first
fringe
away
from
the
maximum,
we
haveì 10- 7 m = 782 nm.
2.60
ì 10
m )away
sin (17.5
è) =
l , which
gives l we
= 7.82
(1central
)central
(the
--66
9
9
3.20 ìì 10
10 m
m))sin
sinqq11 ==((3232))((520
520 ìì 10
10 m
m)),, which
which gives
gives qq11 ==14.1
14.1èè..
((3.20
19. AFor
constructive
interference
from
the
singleby
slit,
the path difference
is is the width of the central
We
find
the
on
from
We
find slit,
the distance
distance
on the
the screen
screen
from
3.
single
1.0mm
wide,
is
illuminated
a 450-nm
light. What
1
D
sin
q
=
m
+
l
,
m
=
1,
2,
3,
...
.
(
)
yy == LLtan
==((10.0
m
2.51m.
maximum
(in
in
the
210.0
tanqqcm)
m)diffraction
tan14.1è=
è= pattern
2.51m. on a screen 5.0m away?
)tan14.1
11
11
For the first fringe away from the central maximum, we have
[3 Marks]
-6
-9
3
20.
We
find
the
angle
to
the
first
minimum
from
3.20
ì
10
m
sin
q
=
520
ì
10
m
,
which
gives
q
=
14.1
è
.
20. We (find the angle )to the1 first
from)
( 2 )minimum
(
1
Chapter 24
The Wave Nature of Light
--99
1
450
ì
10
m
((1)the
ì 10 from
m))
)(( 450
We find the distance
screen
m
mll on
sin
==
==0.00045.
sinqq1min ==
0.00045.
-3
y1 = L1min
tan q1 D
=D(10.0 m
è=
1.0
ìì 10
) tan14.1
1.0
10- 3 m
m 2.51m.
Thus
the width
of the central
maximum is
We
We find
find the
the distance
distance on
on the
the screen
screen from
from
2
y
=
0.0045m
=
0.45cm.
qq.. to the first minimum from
yy ==the
LLtan
tan
20. We find
angle
For
For small
small angles,
angles,
we have
1) ( 450 ì 10- 9 m )
(have
mlwe
sin
= qthe
=
0.00045.
21. The angle
from
centralgives
maximum
to=the
first bright fringe is 16è.
öö tan
sinqq1min
tan
q,, which
which
gives
Dinterference
1.0 ì from
10- 3 m
For constructive
the
single
slit, the path difference is
yy == LLsin
sinqq ==((5.0
5.0m
m)(
0.00045)) ==0.00225m.
0.00225m.
)(0.00045
1 on the screen from
We find
the
distance
D sin q = ( m + 2 ) l , m = 1, 2, 3, ... .
y = L tan q .
For the
firstisfringe
away from
the central
maximum,
we have
4. How
wide
the we
central
diffraction
peak
on a screen
2.30m behind a 0.0348mm wide slit
For
small
angles,Inc.,
haveSaddle
©
Pearson
Education,
NJ.
isis protected
copyright laws
9
©2005
2005
Pearson
Education,
Inc.,
Upper
Saddle-River,
River,
NJ. All
Allrights
rightsreserved.
reserved. This
Thismaterial
material
protected under
under all
laws as
as they
they
3 Upper
D
sin
16
è
=
653
ì
10
m
,
which
gives
D
=
3.6
ì means,
10- 6 without
m
= 3.6
m
m. allincopyright
illuminated
by
a
589nm
light?
(
)
(
)
(
)
currently
material
may
writing
2which
sinNo
tan qofof, this
q portion
öportion
gives
currentlyexist.
exist.
No
this
material
maybe
bereproduced,
reproduced,in
inany
anyform
formor
or by
byany
anymeans,
without permission
permission in
writing from
fromthe
the publisher.
publisher.
209
[3 Marks]
209
y = L sin q = ( 5.0 m )( 0.00045 ) = 0.00225m.
22. We find the angle to the first minimum from
(1) ( 589 ì 10- 9 m )
ml
sin
=
=
= 0.0169, so q This =
0.970isèprotected
.
q
© 2005 Pearson Education,
Inc., Upper Saddle River,
material
under all copyright laws as they
1min
- 3NJ. All rights reserved. 1min
0.0348
ì be
10reproduced,
m)
currently exist. No portion D
of this (material
may
in any form or by any means, without permission in writing from the publisher.
209
We find the distance on the screen from
-2
y1 = L tan q1 = ( 2.30 m ) tan 0.970è= 3.89 ì 10 m = 3.89cm.
Thus the width of the peak is
Dy1 = 2 y1 = 2 ( 3.89cm ) = 7.79cm.
23. We find the angular half-width q of the central maximum from
sin q =
l
;
D
-9
≈55.0è’ 440 ì 10 m
sin ∆
=
, which gives D = 9.53 ì 10- 7 m.
÷
2
D
«
◊
observation
change,
madefarther
at the same
diagram
willmust
be larger,
andbut
so all
theobservations
points C andcould
D willbemove
apart.position.
12. So-called
If the frequency
of thereduction
speakers isdevices
lowered,
then
be increased.
Each
circle inare
the
13.
active noise
work
onthe
thewavelength
principle ofwill
interference.
If the
electronics
diagram
willtobedetect
larger,
so the
points
C and
D will
farther
apart.
fast
enough
theand
noise,
invert
it, and
create
themove
opposite
wave
(180o out of phase with the
original) in significantly less time than one period of the components of the noise, then the original
13. noise
So-called
active
noisenoise
reduction
devices
work on the
of interference.
If the electronics
and the
created
will be
approximately
in aprinciple
destructive
interference relationship.
The are
o
fast
enough
to
detect
the
noise,
invert
it,
and
create
the
opposite
wave
(180
out
of
phase
with
the
person
wearing
the
headphones
will
hear
a
net
sound
signal
that
is
very
low
in
intensity.
Conceptual Questions
original)
in
significantly
less
time
than
one
period
of
the
components
of
the
noise,
then
the
original
(These questions are not in an IB style but instead designed to check your understanding of the concept of this topic. You should
noise the
andtwo
the created
noisebest
will
approximately
in a destructive
interference
Thebeats
14. From
waves
it to
isbe
seen
that the communicate
frequency
of your
beating
is higher
inrelationship.
Figure (a) – the
tryshown,
your
appropriately
answer
using prose)
personmore
wearing
the headphones
hear a net
sound
signal that
is very
intensity.
occur
frequently.
The beatwill
frequency
is the
difference
between
thelow
twoincomponent
frequencies, and so since (a) has a higher beat frequency, the component frequencies are further apart
1.
there
Doppler
if the
andthe
observer
move
in theissame
same
14. Isin
From
wavesshift
shown,
it issource
seen that
frequency
of beating
higherdirection,
in Figure with
(a) – the
the beats
(a). thea two
velocity?
Explain.
occur more frequently. The beat frequency is the difference between the two component
frequencies,
and so since
hassource
a higher
frequency,
component
frequencies
15. There
is no Doppler
shift (a)
if the
andbeat
observer
move the
in the
same direction,
with are
the further
same apart
in (a).
velocity.
Doppler shift is caused by relative motion between source and observer, and if both source
and
observer
move in the same direction with the same velocity, there is no relative motion.
Chapter 12
Sound
15. There is no Doppler shift if the source and observer move in the same direction, with the same
velocity.
shiftbut
is caused
by relative
motion
betweentosource
and observer,
andwill
if both
source
16. If
the windDoppler
is blowing
the listener
is at rest
with respect
the source,
the listener
not hear
a
2. IfDoppler
theobserver
wind
is move
blowing,
will
this
alterofthe
frequency
of
the
sound
heard
by a person
at rest
with
and
in
the
same
direction
with
the
same
velocity,
there
is
no
relative
motion.
effect.
We
analyze
the
case
the
wind
blowing
from
the
source
towards
the
listener.
The
amplitude
(anti-node).
Thus
thewavelength
interferenceof
can
be described
as “interference in time”. To
respect
source?the
Is same
the
velocity
changed?
moving to
airthe
(wind)
effect as if the
speed
of sound
had beeninterference,
increased bytheantime
amount
experience
the
full has
range from
constructive
interference
to destructive
of
16. observation
If the to
wind
blowing
but
the
listener
is
at
rest
with
respect
to
the
source,
the
listener
will
not
hear a
equal
theismust
wind
speed.
The
wavelength
of
the
sound
waves
(distance
that
a
wave
travels
during
change, but all observations could be made at the same position.
Doppler
effect.
Wewill
analyze
the case by
of the
from
thethe
source
listener.
The
one
period
of time)
be increased
the wind
same blowing
percentage
that
wind towards
speed isthe
relative
to the
moving
air
(wind)
has
the
same
effect
as
if
the
speed
of
sound
had
been
increased
by
an
amount
still-air
speed
of
sound.
Since
the
frequency
is
the
speed
divided
by
the
wavelength,
the
frequency
12. If the frequency of the speakers is lowered, then the wavelength will be increased. Each circle in the
equalnot
to the
wind
speed.
The
wavelength
theDsound
waves
(distance
that
wavehas
travels
during
does
change,
and
soand
there
Doppler
effect
to hear.
Alternatively,
thea wind
the same
diagram
will
be larger,
so is
thenopoints
Cofand
will
move
farther
apart.
one period
of time)
willnot
be moving
increasedbut
bythe
thesource
same percentage
themoving
wind speed
relative
to in
thethe
effect
as if the
air were
and listenerthat
were
at theissame
speed
still-air
speed
of
sound.
Since
the
frequency
is
the
speed
divided
by
the
wavelength,
the
frequency
same
direction.
See
question
15
for
a
discussion
of
that
situation.
13. So-called active noise reduction devices work on the principle of interference. If the electronics are
o wind has the same
doesenough
not change,
and the
so there
no Doppler
Alternatively,
fast
to detect
noise,isinvert
it, and effect
createtothehear.
opposite
wave (180the
out of phase with the
effecthighest
as if
the
air
were
not
moving
but
the
source
and
listener
were
moving
atswinging
the then
samethe
speed
in the
17. original)
The
frequency
of
sound
will
be
heard
at
position
C,
while
the
child
is
forward.
in significantly less time than one period of the components of the noise,
original
same
direction.
See
question
15
for
a
discussion
of
that
situation.
Assuming
the
child
is
moving
with
SHM,
then
the
highest
speed
is
at
the
equilibrium
point,
point
noise and the created noise will be approximately in a destructive interference relationship. The C.
And to wearing
have an increased
pitch, will
the relative
motion
the source
detector
be towards each
person
the headphones
hear a net
soundofsignal
that isand
very
low inmust
intensity.
17.
The
highest
frequency
of
sound
will
be
heard
at
position
C,
while
the
child
is
swinging
other.
The child
would
also hear
the lowest
of sound
C, while
swinging
3. The
diagram
shows
various
positions
of afrequency
child in motion
onatapoint
swing.
A monitor
isforward.
blowing a whistle
Assuming
the child isshown,
moving is
with
SHM,
thenfrequency
the highest
speed
is is
at higher
the equilibrium
point,
point
C.
backwards.
14. inFrom
seen
that
the
ofA
beating
Figure
– the
beats
fronttheoftwo
thewaves
child on the itground.
At which
position,
through
E, willinthe
child(a)
hear
the
highest
And tomore
havefrequently.
an increased
pitch,
relative is
motion
of the source
and detector
must be towards each
occur
The
beatthe
frequency
the difference
between
the two component
frequency
thewould
sound
of the whistle?
Explain your
reasoning.
other. Thefor
child
lowest
of the
sound
at point C,
while swinging
frequencies,
and so
sincealso
(a) hear
has athe
higher
beatfrequency
frequency,
component
frequencies
are further apart
backwards.
in (a). to Problems
Solutions
Topic 9.5 Doppler Effect Problems
15.solving
Therethese
is no problems,
Doppler shift
the source
andalways
observer
move
the same
direction,figures
with the
same We
In
the if
authors
did not
follow
theinrules
of significant
rigidly.
velocity.
Doppler
shift
is
caused
by
relative
motion
between
source
and
observer,
and
if
both
source
Solutions
to
Problems
tended to take quoted frequencies as correct to the number of digits shown, especially where other values
and
observer
move
in
the
same
direction
with
the
same
velocity,
there
is
no
relative
motion.
might indicate that. For example, in problem 42, values of 350 Hz and 355 Hz are used. We took both of
In solving
these
problems,
the authors
did not
the rules
of similarly.
significantFor
figures
rigidly.
those
values
to have
3 significant
figures.
We always
treated follow
the decibel
values
example,
in We
16.
If the
wind
is blowing
but theofas
listener
restnumber
with
respect
to the
source,
the listener
willother
not hear
a
tended
to11,
take
frequencies
correct
toathaving
the
of
digits
shown,
especially
where
values
problem
wequoted
treated
the value
120
dBisas
three
significant
figures.
We example,
analyze the
case of the
blowing
the 355
source
listener.
Theof
mightDoppler
indicateeffect.
that. For
in problem
42,wind
values
of 350from
Hz and
Hz towards
are used.theWe
took both
airhave
(wind)
has
same
as iftreated
thesospeed
of sound
had been
increased
by
anofamount
thosemoving
values
to
3 significant
figures.
We
thetime
decibel
values
example,
1.
The
round
trip
time
forthe
sound
is effect
2.0 seconds,
the
for sound
tosimilarly.
travel
theFor
length
the in
lake is
equal
to
the
wind
speed.
The
wavelength
of
the
sound
waves
(distance
that
a
wave
travels
during
problem
11, we treated
thetime
valueand
of the
120speed
dB asofhaving
significant
1.0 seconds.
Use the
soundthree
to determine
thefigures.
length of the lake.
one period of time) will be increased by the same2 percentage that the wind speed is relative to the
d speed
vt of
343m
s 1.0
s the
343 m 3.4 is10themspeed divided by the wavelength, the frequency
sound.
Since
1. still-air
The round
trip time
for sound
is 2.0frequency
seconds, so the time for sound to travel the length of the lake is
notEducation,
change,
and
so
there
isthe
noNJ.
Doppler
effect
hear.
Alternatively,
the
wind
haslaws
theassame
1.0
seconds.
Use
the
time
andRiver,
speed
of sound
to This
determine
length
ofallthe
lake.
© 2005does
Pearson
Inc.,
Upper
Saddle
All rights
reserved.
material
isthe
protected
under
copyright
they
currently
exist. as
Noif
portion
of this
material
be reproduced,
any form
or by any means, without permission in writing from the
effect
the air
were
notmay
moving
but thein source
2 and listener were moving at the same speed in the
publisher.
vt 343m
s 1.0 s 15 343
3.4 10 of
m that situation.
same ddirection.
See question
for amdiscussion
294
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently
exist.
No portion
of this material
may be will
reproduced,
in anyat
form
or by anyC,
means,
without
writing from
the
17. The
highest
frequency
of sound
be heard
position
while
thepermission
child is in
swinging
forward.
publisher.
Assuming the child is moving with SHM, then294
the highest speed is at the equilibrium point, point C.
And to have an increased pitch, the relative motion of the source and detector must be towards each
other. The child would also hear the lowest frequency of sound at point C, while swinging
backwards.
Calculation-based
Questions
Solutions to Problems
In solving
these problems,
the authors
not always
follow thesiren
rulesisof1550Hz
significant
figures
rigidly.
We
1.
The predominant
frequency
of adidcertain
fire engine’s
when
at rest.
What
frequency
tended to take quoted frequencies as correct to the number of digits shown, especially where other values
do you detect if you move with a speed of 30.0m/s (a) toward the fire engine, and (b) away from it?
might indicate that. For example, in problem 42, values of 350 Hz and 355 Hz are used. We took both of
[3 Marks]
those values to have 3 significant figures. We treated the decibel values similarly. For example, in
problem 11, we treated the value of 120 dB as having three significant figures.
1.
The round trip time for sound is 2.0 seconds, so the time for sound to travel the length of the lake is
f beat
f1
f2
v
v
1
2
343 m s
1
1
2.64 m
2.76 m
5.649
5.6 Hz
49. (a) Observer moving towards stationary source.
Chapter 12
Sound
v
30.0 m s
f
1 obs f
1
1550 Hz
1690 Hz
vsnd
343 m s
Observer
moving
away calculate
from stationary
source. of each sound, and then subtract the two
48. (b)
To find
the beat
frequency,
the frequency
frequencies.
v
30.0 m s
f
1 obs f
1
1550 Hz
1410 Hz
v
v
1
1
vsnd
343 m s
f beat
f1 f 2
343 m s
5.649 5.6 Hz
2.64 m 2.76 m
1
2
50. (a) Source moving towards stationary observer.
Hz source.
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Chapter 12
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(b)
moving
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stationary
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1550
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s you detect if a fire engine whose siren emits at 1550Hz
2.moving
You are
standing
still.
Whatthe
frequency
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Topic 9 (New) [105 marks]
1a. Monochromatic light from two identical lamps arrives on a screen.
[1 mark]
The intensity of light on the screen from each lamp separately is I0.
On the axes, sketch a graph to show the variation with distance x on the screen of the intensity I
of light on the screen.
Markscheme
horizontal straight line through I = 2
Accept a curve that falls from I = 2 as distance increases from centre but not if it falls to
zero.
[1 mark]
Monochromatic light from a single source is incident on two thin, parallel slits.
1b. Monochromatic light from a single source is incident on two thin, parallel slits.
[3 marks]
The following data are available.
Slit separation
= 0.12 mm
Wavelength
= 680 nm
Distance to screen = 3.5 m
The intensity I of light at the screen from each slit separately is I0. Sketch, on the axes, a graph
to show the variation with distance x on the screen of the intensity of light on the screen for this
arrangement.
Markscheme
«standard two slit pattern»
general shape with a maximum at x = 0
maxima at 4I0
maxima separated by «
Dλ
=» 2.0 cm
s
Accept single slit modulated pattern provided central maximum is at 4. ie height of peaks
decrease as they go away from central maximum. Peaks must be of the same width
[3 marks]
1c. The slit separation is increased. Outline one change observed on the screen.
[1 mark]
Markscheme
fringe width/separation decreases
OR
more maxima seen
[1 mark]
There is a proposal to power a space satellite X as it orbits the Earth. In this model, X is
connected by an electronically-conducting cable to another smaller satellite Y.
2a. Satellite X orbits 6600 km from the centre of the Earth.
Mass of the Earth = 6.0 x 10 24 kg
Show that the orbital speed of satellite X is about 8 km s –1.
[2 marks]
Markscheme
«v = √
GM E
»
r
= √ 6.67×10
−11
×6.0×1024
6600×103
7800 «m s –1»
Full substitution required
Must see 2+ significant figures.
Satellite Y orbits closer to the centre of Earth than satellite X. Outline why
2b. the orbital times for X and Y are different.
[1 mark]
Markscheme
Y has smaller orbit/orbital speed is greater so time period is less
Allow answer from appropriate equation
Allow converse argument for X
2c. satellite Y requires a propulsion system.
[2 marks]
Markscheme
to stop Y from getting ahead
to remain stationary with respect to X
otherwise will add tension to cable/damage satellite/pull X out of its orbit
2d. The cable between the satellites cuts the magnetic field lines of the Earth at right
angles.
Explain why satellite X becomes positively charged.
[3 marks]
Markscheme
cable is a conductor and contains electrons
electrons/charges experience a force when moving in a magnetic field
use of a suitable hand rule to show that satellite Y becomes negative «so X becomes
positive»
Alternative 2
cable is a conductor
so current will flow by induction flow when it moves through a B field
use of a suitable hand rule to show current to right so «X becomes positive»
Marks should be awarded from either one alternative or the other.
Do not allow discussion of positive charges moving towards X
2e. Satellite X must release ions into the space between the satellites. Explain why
the current in the cable will become zero unless there is a method for transferring
charge from X to Y.
[3 marks]
Markscheme
electrons would build up at satellite Y/positive charge at X
preventing further charge flow
by electrostatic repulsion
unless a complete circuit exists
2f. The magnetic field strength of the Earth is 31 µT at the orbital radius of the
satellites. The cable is 15 km in length. Calculate the emf induced in the cable.
Markscheme
«ε = Blv =» 31 x 10 –6 x 7990 x 15000
3600 «V»
Allow 3700 «V» from v = 8000 m s –1.
The cable acts as a spring. Satellite Y has a mass m of 3.5 x 10 2 kg. Under certain
[2 marks]
The cable acts as a spring. Satellite Y has a mass m of 3.5 x 10 2 kg. Under certain
circumstances, satellite Y will perform simple harmonic motion (SHM) with a period T of 5.2 s.
2g. Estimate the value of k in the following expression.
[3 marks]
T = 2π√ m
k
Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation of
satellite X.
Markscheme
2
use of k = « 4πT 2m =»
4×π2×350
5.22
510
N m–1 or kg s–2
Allow MP1 and MP2 for a bald correct answer
Allow 500
Allow N/m etc.
2h. Describe the energy changes in the satellite Y-cable system during one cycle of the
oscillation.
[2 marks]
Markscheme
Ep in the cable/system transfers to Ek of Y
and back again twice in each cycle
Exclusive use of gravitational potential energy negates MP1
Yellow light from a sodium lamp of wavelength 590 nm is incident at normal incidence on a
double slit. The resulting interference pattern is observed on a screen. The intensity of the
pattern on the screen is shown.
3a. Explain why zero intensity is observed at position A.
[2 marks]
Markscheme
the diagram shows the combined effect of «single slit» diffraction and «double slit»
interference
recognition that there is a minimum of the single slit pattern
OR
a missing maximum of the double slit pattern at A
waves «from the single slit» are in antiphase/cancel/have a path difference of ( n +
)λ/destructive interference at A
1
2
3b. The distance from the centre of the pattern to A is 4.1 x 10 –2 m. The distance from the [2 marks]
screen to the slits is 7.0 m.
Calculate the width of each slit.
Markscheme
θ=
4.1×10−2
7.0
OR b =
λ
θ
«=
7.0×5.9×10−7
»
4.1×10−2
1.0 × 10 –4 «m»
Award [0] for use of double slit formula (which gives the correct answer so do not award
BCA)
Allow use of sin or tan for small angles
3c. Calculate the separation of the two slits.
[2 marks]
Markscheme
use of s =
λD
d
with 3 fringes «
590×10−9×7.0
»
4.1×10−2
3.0 x 10 –4 «m»
Allow ECF.
The double slit is replaced by a diffraction grating that has 600 lines per millimetre. The resulting
The double slit is replaced by a diffraction grating that has 600 lines per millimetre. The resulting
pattern on the screen is shown.
3d. State and explain the differences between the pattern on the screen due to the grating [3 marks]
and the pattern due to the double slit.
Markscheme
fringes are further apart because the separation of slits is «much» less
intensity does not change «significantly» across the pattern or diffraction envelope is
broader because slits are «much» narrower
the fringes are narrower/sharper because the region/area of constructive interference is
smaller/there are more slits
intensity of peaks has increased because more light can pass through
Award [1 max] for stating one or more differences with no explanation
Award [2 max] for stating one difference with its explanation
Award [MP3] for a second difference with its explanation
Allow “peaks” for “fringes”
3e. The yellow light is made from two very similar wavelengths that produce two lines in [3 marks]
the spectrum of sodium. The wavelengths are 588.995 nm and 589.592 nm. These
two lines can just be resolved in the second-order spectrum of this diffraction grating. Determine
the beam width of the light incident on the diffraction grating.
Markscheme
Δλ = 589.592 – 588.995
OR
Δλ = 0.597 «nm»
λ
N = « mΔλ
=»
589
2×0.597
«493»
beam width = « 493
=» 8.2 x 10 –4 «m» or 0.82 «mm»
600
A student is investigating a method to measure the mass of a wooden block by timing the period
A student is investigating a method to measure the mass of a wooden block by timing the period
of its oscillations on a spring.
4a. Describe the conditions required for an object to perform simple harmonic motion
(SHM).
[2 marks]
Markscheme
acceleration/restoring force is proportional to displacement
and in the opposite direction/directed towards equilibrium
A 0.52 kg mass performs simple harmonic motion with a period of 0.86 s when attached to the
spring. A wooden block attached to the same spring oscillates with a period of 0.74 s.
4b. Calculate the mass of the wooden block.
[2 marks]
Markscheme
ALTERNATIVE 1
T12
T22
=
m1
m2
mass = 0.38 / 0.39 «kg»
ALTERNATIVE 2
«use of T
= 2π√ mk » k = 28 «Nm–1»
«use of T
= 2π√ mk » m = 0.38 / 0.39 «kg»
Allow ECF from MP1.
4c. In carrying out the experiment the student displaced the block horizontally by 4.8 cm
from the equilibrium position. Determine the total energy in the oscillation of the
wooden block.
[3 marks]
Markscheme
2π
ω = « 0.74
» = 8.5 «rads–1»
total energy =
1
2
× 0.39 × 8.52 × (4.8 × 10−2 )2
= 0.032 «J»
Allow ECF from (b) and incorrect ω.
Allow answer using k from part (b).
4d. A second identical spring is placed in parallel and the experiment in (b) is repeated.
Suggest how this change affects the fractional uncertainty in the mass of the block.
[3 marks]
Markscheme
spring constant/k/stiffness would increase
T would be smaller
fractional uncertainty in T would be greater, so fractional uncertainty of mass of block would
be greater
With the block stationary a longitudinal wave is made to travel through the original spring from
left to right. The diagram shows the variation with distance x of the displacement y of the coils of
the spring at an instant of time.
A point on the graph has been labelled that represents a point P on the spring.
4e. State the direction of motion of P on the spring.
[1 mark]
Markscheme
left
4f. Explain whether P is at the centre of a compression or the centre of a rarefaction.
[2 marks]
Markscheme
coils to the right of P move right and the coils to the left move left
hence P at centre of rarefaction
Do not allow a bald statement of rarefaction or answers that don’t include reference to the
movement of coils.
Allow ECF from MP1 if the movement of the coils imply a compression.
5a. Outline the conditions necessary for simple harmonic motion (SHM) to occur.
[2 marks]
Markscheme
force/acceleration proportional to displacement «from equilibrium position»
and directed towards equilibrium position/point
OR
and directed in opposite direction to the displacement from equilibrium position/point
Do not award marks for stating the defining equation for SHM.
Award [1 max] for a ω–= 2x with a and x defined.
A buoy, floating in a vertical tube, generates energy from the movement of water waves on the
surface of the sea. When the buoy moves up, a cable turns a generator on the sea bed
producing power. When the buoy moves down, the cable is wound in by a mechanism in the
generator and no power is produced.
The motion of the buoy can be assumed to be simple harmonic.
5b. A wave of amplitude 4.3 m and wavelength 35 m, moves with a speed of 3.4 m s
Calculate the maximum vertical speed of the buoy.
–1.
[3 marks]
Markscheme
frequency of buoy movement =
3.4
35
or 0.097 «Hz»
OR
time period of buoy =
v=«
2πx0
T
35
3.4
or 2πfx0 » =
or 10.3 «s» or 10 «s»
2×π×4.3
10.3
or 2 × π × 0.097 × 4.3
2.6 «m s –1»
5c. Sketch a graph to show the variation with time of the generator output power. Label the[2 marks]
time axis with a suitable scale.
Markscheme
peaks separated by gaps equal to width of each pulse «shape of peak roughly as shown»
one cycle taking 10 s shown on graph
Judge by eye.
Do not accept cos 2 or sin 2 graph
At least two peaks needed.
Do not allow square waves or asymmetrical shapes.
Allow ECF from (b)(i) value of period if calculated.
Water can be used in other ways to generate energy.
Water can be used in other ways to generate energy.
5d. Outline, with reference to energy changes, the operation of a pumped
storage hydroelectric system.
[2 marks]
Markscheme
PE of water is converted to KE of moving water/turbine to electrical energy «in
generator/turbine/dynamo»
idea of pumped storage, ie: pump water back during night/when energy cheap to buy/when
energy not in demand/when there is a surplus of energy
5e. The water in a particular pumped storage hydroelectric system falls a vertical distance [2 marks]
of 270 m to the turbines. Calculate the speed at which water arrives at the turbines.
Assume that there is no energy loss in the system.
Markscheme
specific energy available = «gh =» 9.81 x 270 «= 2650J kg –1»
OR
mgh =
1
mv2
2
OR
v2 = 2gh
v = 73 «ms –1»
Do not allow 72 as round from 72.8
5f. The hydroelectric system has four 250 MW generators. Determine the maximum time [2 marks]
for which the hydroelectric system can maintain full output when a mass of 1.5 x 1010
kg of water passes through the turbines.
Markscheme
total energy = «mgh = 1.5 x 10 10 x 9.81 x 270=» 4.0 x 10 13 «J»
OR
total energy = « 12 mv2 =
time = «
4.0×1013
»
4×2.5×108
1
2
× 1.5 × 1010 × (answer (c)(ii)) 2 =» 4.0 x 10 13 «J»
11.1h or 4.0 x 10 4 s
Use of 3.97 x 10 13 «J» gives 11 h.
For MP2 the unit must be present.
5g. Not all the stored energy can be retrieved because of energy losses in the
system. Explain two such losses.
[2 marks]
Markscheme
friction/resistive losses in pipe/fluid resistance/turbulence/turbine or generator «bearings»
OR
sound energy losses from turbine/water in pipe
thermal energy/heat losses in wires/components
water requires kinetic energy to leave system so not all can be transferred
Must see “seat of friction” to award the mark.
Do not allow “friction” bald.
A student investigates how light can be used to measure the speed of a toy train.
A student investigates how light can be used to measure the speed of a toy train.
Light from a laser is incident on a double slit. The light from the slits is detected by a light
sensor attached to the train.
The graph shows the variation with time of the output voltage from the light sensor as the train
moves parallel to the slits. The output voltage is proportional to the intensity of light incident on
the sensor.
6a. Explain, with reference to the light passing through the slits, why a series of voltage
peaks occurs.
[3 marks]
Markscheme
«light» superposes/interferes
pattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»
voltage peaks correspond to interference maxima
6b. The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10 –7 m. [1 mark]
The slits are 5.0 m from the train track. Calculate the separation between two adjacent
positions of the train when the output voltage is at a maximum.
Markscheme
«s =
λD
d
=
6.3×10−7×5.0
1.5×10−3
=» 2.1 x 10 –3 «m»
If no unit assume m.
Correct answer only.
6c. Estimate the speed of the train.
[2 marks]
Markscheme
correct read-off from graph of 25 m s
v = « xt =
2.1×10−3
25×10−3
=» 8.4 x 10 –2 «m s –1»
Allow ECF from (b)(i)
A student investigates how light can be used to measure the speed of a toy train.
Light from a laser is incident on a double slit. The light from the slits is detected by a light
sensor attached to the train.
The graph shows the variation with time of the output voltage from the light sensor as the train
moves parallel to the slits. The output voltage is proportional to the intensity of light incident on
the sensor.
As the train continues to move, the first diffraction minimum is observed when the light sensor is
at a distance of 0.13 m from the centre of the fringe pattern.
6d. Determine the width of one of the slits.
[2 marks]
Markscheme
angular width of diffraction minimum =
slit width = « λd =
6.3×10−7
0.026
0.13
5.0
«= 0.026 rad»
=» 2.4 x 10 –5 «m»
Award [1 max] for solution using 1.22 factor.
6e. Suggest the variation in the output voltage from the light sensor that will be observed [2 marks]
as the train moves beyond the first diffraction minimum.
Markscheme
«beyond the first diffraction minimum» average voltage is smaller
«voltage minimum» spacing is «approximately» same
OR
rate of variation of voltage is unchanged
OWTTE
6f. In another experiment the student replaces the light sensor with a sound sensor. The [2 marks]
train travels away from a loudspeaker that is emitting sound waves of
constant amplitude and frequency towards a reflecting barrier.
The graph shows the variation with time of the output voltage from the sounds sensor.
Explain how this effect arises.
Markscheme
«reflection at barrier» leads to two waves travelling in opposite directions
mention of formation of standing wave
maximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position
Police use radar to detect speeding cars. A police officer stands at the side of the road
7a. Police use radar to detect speeding cars. A police officer stands at the side of the road [6 marks]
and points a radar device at an approaching car. The device emits microwaves which reflect off
the car and return to the device. A change in frequency between the emitted and received
microwaves is measured at the radar device.
The frequency change Δf is given by
Δf =
2fv
c
where f is the transmitter frequency, v is the speed of the car and c is the wave speed.
The following data are available.
Transmitter frequency f = 40 GHz Δf = 9.5 kHz Maximum speed allowed = 28 m s–1
(i) Explain the reason for the frequency change.
(ii) Suggest why there is a factor of 2 in the frequency-change equation.
(iii) Determine whether the speed of the car is below the maximum speed allowed.
Markscheme
i
mention of Doppler effect
OR
«relative» motion between source and observer produces frequency/wavelength change
Accept answers which refer to a double frequency shift.
Award [0] if there is any suggestion that the wave speed is changed in the process.
the reflected waves come from an approaching “source”
OR
the incident waves strike an approaching “observer”
increased frequency received «by the device or by the car»
ii
the car is a moving “observer” and then a moving “source”, so the Doppler effect occurs
twice
OR
the reflected radar appears to come from a “virtual image” of the device travelling at 2 v
towards the device
iii
ALTERNATIVE 1
For both alternatives, allow ecf to conclusion if v OR Δf are incorrectly calculated.
v=«
(3×108)×(9.5×103)
(40×109)×2
=» 36 «ms –1»
«36> 28» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.
ALTERNATIVE 2
reverse argument using speed limit.
9
Δf = « 2×40×108×28 =» 7500 «Hz»
3×10
« 9500> 7500» so car exceeded limit
There must be a sense of a conclusion even if numbers are not quoted.
7b. Airports use radar to track the position of aircraft. The waves are reflected from the
[2 marks]
aircraft and detected by a large circular receiver. The receiver must be able to resolve
the radar images of two aircraft flying close to each other.
The following data are available.
Diameter of circular radar receiver = 9.3 m Wavelength of radar = 2.5 cm Distance of two
aircraft from the airport = 31 km
Calculate the minimum distance between the two aircraft when their images can just be
resolved.
Markscheme
x=
31×103×1.22×2.5×10−2
9.3
Award [2] for a bald correct answer.
Award [1 max] for POT error.
100 «m»
Award [1 max] for 83m (omits 1.22).
Monochromatic light is incident normally on four thin, parallel, rectangular slits.
The graph shows the variation with diffraction angle θ of the intensity of light I at a distant
screen.
I0 is the intensity of the light at the middle of the screen from one slit.
8a. Explain why the intensity of light at θ=0 is 16I0.
[3 marks]
Markscheme
constructive interference
amplitude/amount of light from 4 slits is 4 × amplitude «from one slit»
intensity is proportional to amplitude 2 OR shows 4 2 = 16 in context of intensity
8b. The width of each slit is 1.0µm. Use the graph to
[4 marks]
(i) estimate the wavelength of light.
(ii) determine the separation of two consecutive slits.
Markscheme
(i)
«diffraction minimum at» θ=0.43rad
λ =≪ bθ = 1.0 × 10−6 × 0.43 =≫ 4.3 × 10−7 m
Accept θ in range 0.41 to 0.45 rad.
Allow λ=bsinθ but do not allow nλ=dsinθ.
Award [1 max] for solution using factor of 1.22.
Award [0] if use of
s = λD
seen.
d
(ii)
«first secondary maximum at» θ=0.125rad
d=
1×valuefrom(b)(i)
sin 0.125
= 3.4 × 10−6 m
Accept q in range 0.123 to 0.127 rad.
Sine must be seen to award MP2.
Allow ECF from (b)(i).
Allow use of 2nd or 3rd maxima (0.25 rad and 3.46 µm or 0.375 rad and 3.5 µm with
appropriate n).
8c. The arrangement is modified so that the number of slits becomes very large. Their
separation and width stay the same.
[4 marks]
(i) State two changes to the graph on page 20 as a result of these modifications.
(ii) A diffraction grating is used to resolve two lines in the spectrum of sodium in the second
order. The two lines have wavelengths 588.995nm and 589.592nm.
Determine the minimum number of slits in the grating that will enable the two lines to be
resolved.
Markscheme
(i)
primary maxima/fringes become brighter/more intense
primary maxima become narrower/sharper
secondary maxima become unimportant/less intense/disappear
Insist on “secondary” for MP3.
(ii)
N =≪
λ̄
mΔλ
=≫
N=494 or 500
589.2935
2×0.5970
Allow use of 588.995 nm or 589.592 nm for
λ̄.
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.
9a. (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.
[4 marks]
Markscheme
(i)
ALTERNATIVE 1
«distance travelled by wave =» 0.30 m
v =≪
distance
time
=≫ 340ms−1
ALTERNATIVE 2
evaluates T =
0.882×10−3×1.6
«=4.7ms»
0.3
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
ω =≪ 2πf =≫ 2π ×
340
1.60
= 1.3 × 103 or 1.34×10 3rads–1
ALTERNATIVE 2
«0.882 ms is
0.3
1.6
of cycle so whole cycle is»
1.35×103rads–1
2π×3
16×0.882×10−3
Allow ECF from (b)(i).
9b. One particle in the medium has its equilibrium position at x=1.00 m.
[4 marks]
(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.
Markscheme
(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 = 2π × 212 × √(4.0 × 10−3 ) − (1.5 × 10−3 )
2
2
«v=4.939≈4.9ms-1»
Allow ECF from (b)(ii).
Do not allow
4.3mm
0.882ms
= 4.87ms−1 .
The travelling wave in (b) is directed at the open end of a tube of length 1.20 m. The
9c. 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.
[3 marks]
(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)
the superposition/interference of two oppositely moving/reflected «identical travelling»
waves
(ii)
the allowed wavelengths in the tube are λ =
4L
n
=
480
n ,
n = 1, 3, 5,…
OR
diagram showing
1.6 =
4.80
n
3
4
of a standing wavelength in the tube
⇒n=3
OR
justification that
3
4
× 1.6 = 1.2m
Allow diagram showing
3
of a wavelength for MP1.
4
© International Baccalaureate Organization 2019
International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
Printed for Concordian International School
Markscheme-Topic 9: Wave phenomena
1.
B
[1]
2.
D
[1]
3.
C
[1]
4.
A
[1]
5.
C
[1]
6.
A
[1]
7.
C
[1]
8.
A
[1]
9.
D
[1]
10.
C
[1]
11.
A
[1]
12.
A
[1]
13.
D
[1]
Short answer questions.
14.
Simple harmonic motion and the greenhouse effect
(a)
the force acting / accelerating (on the body) is directed towards equilibrium
(position);
and is proportional to its / the bodies displacement from equilibrium;
(b) (i)
(ii)
1.5 × 1010 m;
2
1
T = 1.1 × 1012 s;


1
;
f  
12 
 1.110 
= 9.1 × 1013 Hz
(iii)
2
ω = (2f) = 5.7 × 1014 (rad s1);
Emax =
 mω x 
1
2
2
2
0
1
2
1.7 10 27  1.5 10 20  5.7 10 28 ;
2
= 6.2 × 1018 J
(c) negative sine;
starting at zero;
with same frequency as displacement; (allow
2
2
2mm square)
3
1
(d)
(i)
k = (42f2mp) = 40 × 83 × 1026 × 1.7 × 1027;
 560 Nm1
(ii)
1
use of F = kx and F = ma;
to give a 
560 1.5 10 10
 5.0 1019 ms 2 ;
 27
1.7 10
2
[13]
15.
(a)
(b)
each element of the slit acts as a point source of light;
the light from these sources interfere;
there will be a zero of intensity (on the screen) when the sum of the
path differences between the sources is an integral number of half
wavelengths / a maximum when an integral number of wavelengths;
3
θ d  λ ;
D b
Dλ ;
rearrange to get d 
2
b
(c)
central maximum same intensity as single slit maximum;
two other maximum either side about half-intensity of central maximum;
Award [1 max] if lines do not touch x-axis.
There is no need to show maxima within secondary maxima. Do not penalize
responses if more than two maxima are shown but they must be symmetrical
and with realistic relative intensities.
2
[7]
16.
(a)
the net displacement of the medium / particles (through which waves travel);
is equal to the sum of individual displacements (produced by each wave); 2 max
Award a good understanding [2 max] and a reasonable one [1 max].
(b) Wave X and wave Y should be identical.
X and Y
A
B
correct phase for wave X;
correct phase for wave Y;
amplitudes the same for each wave;
amplitude for each wave is two divisions;
4 max
2
(c)
(i)
the phase difference between light leaving S1 and S2 is constant;
1
Do not penalize the candidate if they state “has the same phase”.
(ii) to produce sufficient diffraction;
for the beams to overlap;
OWTTE;
2 max
[9]
17.
(a)
Award [2] for a clear statement or [2] for a clear diagram.
the maximum of one diffraction pattern is coincident with the first
minimum of the other;
or:
2
(b)
θ min 1.22
λ
d
(with small angle approximation), θ 
equate θ 1.22
s
;
150 m
590 10 9
λ
1.22
;
D
5.0 10 3
to get s = 2.2 cm;
3
Award [2 max] if 1.22 factor is omitted.
[5]
18.
(a)
(b)
circular wavefronts originating from four successive source positions;
bunching of wavefronts in front, spreading out at back;
approximately, correct spacing of wavefronts in front, and behind source;
3
f waves in distance (V – v);
(V  v)
apparent wavelength =
;
f
apparent frequency =
f V
;
(V  v)
3
Allow any other valid and correct approach or statement of formula.
Award [0] for quote of formula with no working shown.
(V  v)
;
V
600  (3  108  v)
599.996 =
;
(3  108 )
(c) ’ = 
v = 2000 m s–1;
3
Allow alternative version for red-shift.
[9]
3
19. (a)
identification of path length differences from slit to slit;
to give constructive interference at a particular angle for a
particular wavelength;
thus different wavelengths will constructively interfere at different
angles ie light will be separated in component wavelengths;
3 max
Award full marks for other explanations not of this format but the
response must explain the creation of the spectrum.
(b)
correct substitution into nλ = d sin θ;
to give sin θ = 5.896 × 10–7 × 600 000 = 0.35376 so θ = 20.7° 21°;
2
[5]
20. (a)
(b)
same number of maxima at the same place but much sharper;
greater intensity than double slit / presence of small maxima in between maxima; 2

;
d
sin θ =
= 0.36 to give θ = 21°;
2
[4]
21. (a)
(i)
180 degrees (180 ) / ; (no unit required if answer given is  )
1
λ
Accept .
2
(ii)
(b)
none;
1
need one wavelength path difference;
2nt = ;
t = 2.2 × 107 m;
3
If same answer to (a)(i) and (ii) then allow t 
λ
as ecf for full marks.
4
[5]
22. (a)
 / same phase change on reflection at upper and at lower surfaces;
for destructive, path difference must be
1
2
λ;
d λ;
4
(b)
destructive interference for one colour / wavelength / green only;
other colours / red and blue still reflected giving colouring / purple colour;
3
2
[5]
4