Name :_______________________________________
“SOME” REVIEW QUESTIONS AND CONCEPTS FOR “Quiz”
Answers on Answer Sheet
Waves and Oscillations - Topic 4 , Topic 11 HL and Option A SL
Formulas – See Formula Sheet
Definition of a TRAVELLING wave :
a wave is a disturbance that travels in a
medium transferring energy and momentum from one place to another. The direction of energy
transfer is the direction of propagation of the wave.
1
Example 1 :
One end of a long string is vibrated at a constant frequency f. A travelling wave of wavelength 
and speed v is set up on the string.
The frequency of vibration is doubled but the tension in the string is unchanged. Which of
the following shows the wavelength and speed of the new travelling wave? A
Wavelength
A.
B.
λ
2
λ
2
Speed
v
2v
C.
2
v
D.
2
2v
v=λf
wavelength is inversely proportional to f. speed stays the same
2
Energy and the Pendulum
-X max
X=0
+ Xmax
v=0
vmax
+amax
a=0
-amax
+ Fmax
F=0
- Fmax
Ep max
Ep = 0
Ep max
Ek = 0
Ek max
Ek = 0
v=0
3
SHM displacement – amplitude, velocity and acceleration graphs
4
5
Example 2:
A - amax follows xmax BUT OPPOSITE IN DIRECTION
Observe that the slope is negative , –ω2 , and a is opposite in direction to X. In other words,
when a is positive X is negative and vice versa.
6
Example 3 : A wooden block is at rest on a horizontal frictionless surface. A horizontal spring is
attached between the block and a rigid support.
The block is displaced to the right by an amount X and is then released. The period of
oscillations is T and the total energy of the system is E.
For an initial displacement of
X
which of the following shows the best estimate for the
2
period of oscillations and the total energy of the system?
Period
A.
T
B.
T
C.
D.
T
2
T
2
Total energy
E
2
E
4
E
2
E
4
B
The period ( T ) is independent of amplitude (A) or displacement (X). The energy is
proportional to A2 or X2 since Ep = ½ k x2.
T for X or X is the same.
2
Total energy of system E for X = 1 E
If X is decreased by ½ (X ) the total energy is decreased by (1/2)2 = 1/4
2
So total energy of system is = E
4
7
Tsokos p. 211: 16
8
p. 211 16 c
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Doppler Effect
Example 4: A source of sound is moving at a steady speed directly towards an observer. The
sound heard by the observer will
a.
b.
c.
d.
Steadily increase in pitch due to longer wavelength.
Steadily decrease in pitch due to shorter wavelength.
Steadily increase in pitch due to shorter wavelength.
Steadily decrease in pitch due to longer wavelength.
Wave Phenomena : Reflection and Refraction
Open the following link http://www.physicsclassroom.com/Class/waves/
Boundary Behavior
Read Lesson 3a and answer the following questions:
1. Define boundary.
When the incident pulse reaches the boundary, two things occur:
 A portion of the energy carried by the pulse is reflected and returns towards the left end of
the rope. The disturbance which returns to the left after bouncing off the pole is known as the
reflected pulse.
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 A portion of the energy carried by the pulse is transmitted to the pole, causing the pole to
vibrate.
2. Describe and explain the reflected pulse that occurs from fixed end reflection [note: the
animation may be useful].
3. Describe and explain the reflected pulse that occurs from free end reflection [note: the
animation may be useful].
11
Example 5 :
A pulse is sent down a string fixed at one end.
A.
C.
Which one of the following diagrams best represents the reflected pulse? C inverts from
fixed end
B.
D.
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Refraction
When waves strike a transparent surface part of it is reflected and part of it is transmitted where it is
refracted. Refraction is the bending of waves when they pass from one medium to another. It occurs
because the wave speed changes as it passes through different mediums (analogous to driving a car
along the edge of the road: if one tire moves off the edge of the road into sand or thick mud it slows
down, pulling the car off to the side of the road)
In the diagrams below the wave is passing from a less dense medium to a more dense medium and so
the wave slows down and bends or refracts.
ADD: As wave travel to more dens medium wavelength, velocity and amplitude DECREASE
BUT FREQUENCY DOESN’T CHANGE
Normal Line of reference
In diagram (b) the RAY is said to bend toward the normal when going from a less dense to a more
dense material. add: The WAVE bends UP.
In diagram( b )below light is passing from water to air or more dense to less dense and the RAY bends
away from the normal. The WAVE bends DOWN.
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26 – 28 only
14
15
Example 6 : A plane wave approaches and passes through the boundary between two media. The
speed of the wave in medium 1 is greater than that in medium 2. Which one of the following
diagrams correctly shows the wave fronts? A waves bend toward the normal
A.
B.
Medium 1
Medium 1
Medium 2
Medium 2
C.
D.
Medium 1
Medium 2
Medium 1
Medium 2
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Standing Waves - Tube Open at Both Ends – Resonance
Wind Instruments
Standing waves can be established in a tube that is open at both ends because the sound
waves can be reflected by the open air. The tube length can be any multiple of ½ λ (i.e. L =
nλ/2; where n = 1, 2, 3, ...). Equations are the same as for standing waves on a string.
2L
n
nv
fn 
2L
n 
(Harmonic wavelengths, tube open at both ends, for any integer n = 1, 2, 3, …;)
(Harmonic frequencies, tube open at both ends, for any integer n = 1, 2, 3, …;)
Example 7:
What will be the fundamental frequency and 2nd and 3rd harmonics for an organ pipe open at
both ends if the length of the pipe is 26 cm and the speed of sound in air is 343ms-1 ?
Remember open at both ends you get all harmonics. Closed at 1 end you only get odd
fn 
nv
2L
1st = fo = 659.6 Hz
2nd = 2 fo = 1319.2 Hz
3rd = 3 fo = 1978.8 Hz
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Standing Waves - Tube Closed at One End – Resonance
Wind Instruments
Resonance – read Tsokos p. 255
Resonance Standing waves can be established in a tube that is closed at one end if the tubes
length is equal to an odd multiple of ¼ λ (i.e. L = nλ/4; where n = 1, 3, 5, …).
n 
4L
n
(Harmonic wavelengths, closed at one end, for any odd integer n = 1, 3, 5,
add: L = nλ
L = nv
4
fn 
nv
4L
4f
(Harmonic frequencies, closed at one end, for any odd integer n = 1, 3, 5,
Figure 12-12
Modes of
vibration
(standing
waves) for a
tube closed
at one end
“closed tube”
Example 8 :
What will be the fundamental frequency and 2nd and 3rd harmonics for an organ pipe closed at
one end if the length of the pipe is 26 cm and the speed of sound in air is 343ms-1 ?
fn 
nv
4L
1st = fo = 659.6 Hz
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3rd = 3 fo = 1978.8 Hz
Example 9 : Standing waves in an open pipe come about as a result of A ( see definition topic
11 option a notes)
A.
reflection and superposition.
B.
reflection and diffraction.
C.
superposition and diffraction.
D.
reflection and refraction.
Example 10 : A pipe, open at one end, has a length L. The speed of sound in the air in the pipe is
v. The frequency of vibration of the fundamental (first harmonic) standing wave that can be set up
in the pipe is D standing waves
nv
4L
fn 
A.
v
.
2L
B.
L
.
2v
C.
4v
.
L
D.
v
4L
.
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Example 11 : A vibrating tuning fork is held above the top of a tube that is filled with water.
The water gradually runs out of the tube until a maximum loudness of sound is heard. C
Maximum loudness occurs at antinode
fn 
A.
nv
4L
= ¼ L
Which of the following best shows the standing wave pattern set up in the tube at this
position?
B.
D.
C.
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Diffraction Tsokos pp. 238 – 240
Tsokos p. 242 : # 3 ( diffraction)
Example 12: Jeremy is walking alongside a building and is approaching a road junction. A fire engine is
sounding its siren and approaching the road along which Jeremy is walking. D Sound bending around
corners is diffraction.
Jeremy
Building
Fire engine
Jeremy cannot see the fire engine but he can hear the siren. This is due mainly to
A.
reflection.
B.
refraction.
C.
the Doppler effect.
D.
diffraction.
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Single Slit Diffraction Tsokos pp. 261 -265 SL Text p. 96 – 97, 226 –
227
A single slit diffraction pattern is produced when light passes through a single aperture
( opening) that is same size as the wavelength of light being used. The light will spread out and
produce a pattern of bright spots called maxima, due to constructive interference , and dark
spots called minima , due to destructive interference, as seen in the figure below.
Example :
A single slit width 0.0015 mm is illuminated with light of wavelength 500.0 nm. Find the angular
width of the central maximum in degrees ( not rads).
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intensity
 
Explain qualitatively, this intensity distribution.
...................................................................................................................................
Resolution and the Rayleigh Criterion SL text pp. 228 – 230, Tsokos
pp. 267-270
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Brewster's Law
In 1812, Sir David Brewster found experimentally that, the reflected ray is 100% polarized
when the angle between the refracted ray and the reflected ray is 90º. The angle of incidence,
called the Brewster's or Polarizing angle θp, required for 100% polarization is determined by the
refractive index of the two materials (using Snell's Law).
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25
MISCELLANEOUS QUESTIONS TOPIC 4 and 11
1.
One end of a horizontal string is fixed to a wall. A transverse pulse moves along the string as
shown.
Which of the following statements are correct for the reflected pulse compared to the
forward pulse? C
I.
It moves more slowly.
II.
It has less energy.
III.
It is inverted.
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
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2.
The two graphs show the variation with time of the individual displacements of two waves as
they pass through the same point. B
The displacement of the resultant wave at the point at time T is equal to
3.
A.
x1 + x2 .
B.
x1 – x2 .
C.
A1 + A2.
D.
A1 – A2.
Which of the following correctly describes the change, if any, in the speed, wavelength and
frequency of a light wave as it passes from air into glass? A
Speed
Wavelength
Frequency
A.
decreases
decreases
unchanged
B.
decreases
unchanged
decreases
C.
unchanged
increases
decreases
D.
increases
increases
unchanged
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ADD: As wave travel to more dens medium wavelength, velocity and amplitude DECREASE BUT
FREQUENCY DOESN’T CHANGE
4.
This question is about simple harmonic motion (SHM) and a wave in a string.
(a)
By reference to simple harmonic motion, state what is meant by amplitude.
Answer : the maximum displacement of the system from equilibrium/
from centre of motion / OWTTE;
1
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(b)
A liquid is contained in a U-tube.
Diagram 1
Diagram 2
The pressure on the liquid in one side of the tube is increased so that the liquid is
displaced as shown in diagram 2. When the pressure is suddenly released the liquid
oscillates.
The damping of the oscillations is small.
(i)
Describe what is meant by damping.
Answer : the amplitude of the oscillations or (total) energy decreases
with time because a force always opposes direction of motion. Force is
a resistive force or frictional force
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(ii)
The displacement of the liquid surface from its equilibrium position is x. The
acceleration a of the liquid in the tube is given by the expression
a= 
2g
x
l
where g is the acceleration of free fall and l is the total length of the liquid
column.
The total length of the liquid column in the tube is 0.32 m. Determine the period
of oscillation.
Answer : T = 0.80 s
ω2 = slope = 2g
l
ω=
2g
;
l
ω=2π
T
T = 2π
0.32
;
2  9.81
T = 0.80 s
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(c)
A wave is travelling along a string. The string can be modelled as a single line of
particles and each particle executes simple harmonic motion. The period of oscillation
of the particles is 0.80 s.
The graph shows the displacement y of part of the string at time t = 0. The distance
along the string is d.
(i)
On the graph, draw an arrow to show the direction of motion of particle P at the
point marked on the string.
Answer :
(ii)
arrow points upward
Show that the speed of the wave is 5.0 m s–1.
λ = 4.0 m from graph
recognition that f = 1/T =
1
= 1.25
0.80
v = f λ = 1.25 × 4.0
v = 5.0 m s–1
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5.
This question is about simple harmonic motion.
A tuning fork is sounded and it is assumed that each tip vibrates with simple harmonic
motion.
The extreme positions of the oscillating tip of one fork are separated by a distance d.
A) State, in terms of d, the amplitude of vibration.
b) The sketch graph below shows how the velocity of a tip varies with time.
On the axes, sketch a graph to show how the acceleration of the tip varies with time.
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Answer:
remember:
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6)
In a simple model of a methane molecule, a hydrogen atom and the carbon atom can
be regarded as two masses attached by a spring. A hydrogen atom is much less
massive than the carbon atom such that any displacement of the carbon atom may be
ignored.
The graph below shows the variation with time t of the displacement x from its
equilibrium position of a hydrogen atom in a molecule of methane.
The mass of hydrogen atom is 1.7  10–27 kg. Use data from the graph above
(i)
to determine its amplitude of oscillation.
Answer = 1.5  1010 m;
(ii)
to show that the frequency of its oscillation is 9.1  1011 Hz.
Answer : f = 9.1  1011 Hz
From graph : T = approximately 0.11 x 10 -13 = 1.1  1012 s
f= 1/T

1
f  
12
 1.110

 ;

= 9.1  1011 Hz
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(iii)
Use the formula below ( see data booklet Topic 4)
to show that the maximum kinetic energy of the hydrogen atom is 6.2  10–22 J.
ω = (2f) or 2 =
T
5.7  1012
m = mass of hydrogen atom = 1.7  10–27 kg
x0 = max displacement = amplitude = 1.5  1010 m
Emax =

1
2
mω 2 x 0
2

1
2
 1.7  10 27  1.5  10 20  5.7   10 28 ;
2
2
= ½ (1.7  10–27 kg) x (5.7  1012 )2 x (1.5  1010 )2
= 6.2  1022 J
(iv)
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7. Jeremy is walking alongside a building and is approaching a road junction. A fire engine is
sounding its siren and approaching the road along which Jeremy is walking. D
Jeremy
Building
Fire engine
Jeremy cannot see the fire engine but he can hear the siren. This is due mainly to
A.
reflection.
B.
refraction.
C.
the Doppler effect.
D.
diffraction.
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