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AS Physics Unit 12
Waves
Mr D Powell
Chapter Map
Don’t be confused by the
necessarily transverse
depiction of (longitudinal)
sound waves on an
oscilloscope.
Differentiating between
wavelength and time
period is very important
here.
Also remember your units.
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Index
12.1 Waves and vibrations
Spec Link 3.2.3: Progressive waves;
Longitudinal and transverse waves
1.
AO1a/AO2b – Explain difference
between transverse and
longitudinal waves? (A-E)
What are the differences between
transverse and longitudinal waves?
A01a – Identify T or L waves. (E)
2.
What is a plane-polarised transverse
wave?
AO1a – Explain how planepolarised transverse waves behave.
(B-E)
3.
What physical test can distinguish
transverse waves from longitudinal
waves?
AO2b – Put polarisation in a real
context such as radio waves, sun
glasses, stress testing, measuring
concentrations (A*-D)
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Longitudinal Waves
Using a slinky you can
try this out.
In fact we are
modelling how the air
molecules compress
and expand when we
talk.
Rarefaction is an
expansion!
Try it with a slinky!
VIBRATION
• Common examples:- Sound, slinky springs
seismic p waves
• Longitudinal waves cannot be polarised
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Longitudinal
Direction of travel
VIBRATION
• The direction of vibration of the particles is parallel to the
direction in which the wave travels.
• Common examples:- Sound, slinky springs
seismic p waves
• Longitudinal waves cannot be polarised
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5
Transverse
Direction of travel
vibration
• The direction of vibration of the particles is perpendicular
to the direction in which the wave travels.
• Common examples:- Water, electromagnetic, ropes, seismic
s waves
• You can prove that you have a transverse wave if you can
polarise the wave (especially important with light (electromagnetic) as you cannot
“see” the wave!!)
Try it with a slinky!
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6
Exam Question…. (Basic Level)
(a) State the characteristic features of
(i)
longitudinal waves,
............................................................................................................................
............................................................................................................................
(ii) transverse waves.
............................................................................................................................
............................................................................................................................
(3)
Answer
a)
(i) particle vibration (or disturbance or oscillation) (1)
same as (or parallel to) direction of propagation (or energy transfer) (1)
(ii)
(particle vibration)
perpendicular to direction of propagation (or energy transfer) (1)
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Polarisation
Create your own diagram
to show this concept
clearly.
Then explain it to another
student.
Electric field vector
As you rotate 90 or
/2 the light
gradually fades
Try it with a polariser!
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Applications of Transverse Polarisation…
Can you research each one of these ideas and see how Polarisation has an
impact. Draw out a mind map and write out the key points for each one. You only
need the basic idea for the exam not the details….
Radio? (GCSE)
Calculator?
(Harder)
Concentration
Transverse
Polarisation
Sun Glasses
G&T Sheet 12_1
Stress Testing
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Uses of Polarisation
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Fishing?
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Using polarisation to measure concentration
1. Some liquids are ‘optically active’ and rotate the electric vector.
2. The liquid’s concentration is proportional to the electric vector
rotation.
laser
Sugar
solution
polariser
©
John Parkinson
analyser
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Index
12
Stress Analysis
The structure of certain plastics will show
polarisation.
When viewed under stress the structure
polarises the light differently.
The place where stress is greatest shows a
more rapid colour change.
Models can be made of complex
components which are viewed with a
polarising filter so engineers can design out
the stresses.
©
John Parkinson
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13
Calculator LCD Displays (Harder)
Nematic Crystals
1.
Polariser filter film with a vertical axis to
polarize light as it enters.
2.
Glass with electrodes to show patterns
when the LCD is turned ON.
3.
Twisted nematic liquid crystal. Rotates light
90 or /2 when turned on.
4.
Glass substrate with electrode film
5.
Polarising filter film with a horizontal axis
to block/pass light.
6.
Reflective surface to send light back to
viewer.
KEY Point. System allows on/off change of transmission by use of
Twisted nematic liquid crystal & crossed polarisers
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Polarisation Exam Question…. (Basic Level)
(b) Daylight passes horizontally through a
fixed polarising filter P. An observer views
the light emerging through a second
polarising filter Q, which may be rotated in
a vertical plane about point X as shown in
the diagram.
Describe what the observer would see as
Q is rotated slowly through 360°. (1 mark)
Answer
(b)
variation in intensity between max and min (or light and dark) (1)
or
two maxima (or two minima) in 360° rotation (1)
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12.2 Measuring waves
Spec.. 3.2.3: Progressive waves;
Longitudinal and transverse waves
1.
2.
3.
What is meant by the amplitude of a
wave?
Between which two points can the
wavelength be measured?
How is the frequency of a wave
calculated from its period?
phase shift
A01a – Be able to explain the
terms: amplitude, frequency,
wavelength, speed, phase (A-E)
AO1a - Be able to: explain path
difference and link to time or (AC)
A02b – Apply knowledge to
oscilloscope questions and lissajou
figures (extension)
2d
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What is missing?
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Measuring Waves
Create a small summary for each of these key terms. Define what is means and
draw a diagram to support your notes…
Use page 176 to assist you…
Displacement
Wave speed
Amplitude
Measuring
Waves
Frequency
Wavelength
Cycle
Wiki
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Key Summary…
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Timings?
Q. A mains transformer vibrates the floor at 50Hz.
What is the time for a complete cycle?
360 o =
2 radians
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Wave Speed Equation
speed = distance
time
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Wave Speed Equation
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Phase Difference
• These two waves are not “coherent”
• What we are saying is all the peaks
don’t match at the same time.
• They have a “phase difference”
360 2
shift 2
• We can express this quite simply as a
2d
time shift or distance shift.
shift
d
d
• If we are talking about a distance
shift then a circle or cycle has 2
radians or 360 to return to the
start.
Shift = 360 * (60mm/240mm)
= 51.4
or T
or
• We can express the shift as a
Shift = 2* (60mm/240mm)
fraction of the circle/ cycle where
= 0.9 radians
is a whole cycle and d is the shift or…
Wiki
NB: it often helps to think of waves as cycles or a clock face and
two hands with a distance (or angle between them)
d
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Phase difference examples…
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Lissajou Figures… (Extension)
If we try using an Oscilloscope to show the idea we can plot two inputs
against each other to form a dynamic graph.
Input 1 = x, Input 2 = y
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b) ¾ of cycle later Q would be at a trough point and
returning to centre
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Plenary Question….
What is the phase shift in degrees
for these waves if…
1) d = 0.1m, = 1.2m
2) d = 22mm, , = 1.2m
3) Harder. Now repeat for radians.
1) Shift = 360 * (0.1m/1.2m)
= 30
d
360 o =
2) Shift = 360 * (22 x 10-3m/1.2m)
= 6.6
360 2
3)
shift 2
= 0.52 radians
= 0.12 radians
shift
2d
2 radians
d
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12.3 Wave properties 1
Specification link-up 3.2.3: Refraction at a plane surface; Diffraction
1. What causes waves to refract when they pass across a boundary?
2. In which direction do light waves bend when they travel out of glass and into
air?
3. What do we mean by diffraction?
Be able to clearly explain the concept and features
of…
A01a –reflection off a plane surface (water / light
waves) (Basic)
A01a –refraction at a boundary (water / light waves
slow in shallow) (Basic)
A01a –diffraction more for narrow gap or longer
(water waves)
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Wave behaviour?
Shouting
around a
corner?
Mirrors
Wavespeed
Fuzzy edges on
a shadow
Wavefronts
Water waves
slow down in
the shallows
Light slows
down in a
medium
Wavelength
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Key Points
Refraction….
Waves pass a boundary i.e. air to
glass prism or deep to shallow water
they “refract” or change direction
and change speed. Light bends in
towards the normal for air to glass
and reverse as it comes out. Water
waves moving into the shallows
slow down and have smaller .
c= f so if c so
Reflection….
Hard surface, angle incident = angle
reflection (from normal). If water waves the
fronts act at 90 to the direction of travel
Diffraction….
Wave fronts incident on a gap. The
narrower the gap the more the waves
curve or the longer/larger the wavelength
the more they spread out.
Video
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Virtual Ripple Tanks…
Use the virtual ripple tank here to explore
wave properties.
http://www.falstad.com/ripple/
Use the ideas from the book on page 179 and
also you can download the additional
information sheet on the blog to help you
explore the ideas.
Make summary notes on what you find for
each situation. You may decide to screenshot
out the image to help you. (NB: pick a nice
colour scheme)
W:\Students Read Only\Physics\AS\Unit 12\ripple tank\index.html
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Plenary Question.
1) Name one example each of reflection,
refraction, and diffraction. (3 marks)
a) Reflection:
b) Refraction:
c)
Diffraction:
2) Then go on to explain at least 1 key
feature of the Physics related to the
process for each. (3marks)
Reflection: Mirror surface, light
travels at constant speed but
changes direction on impact
Refraction: Light enters a glass
prism which is more dense than air
so slows down (smaller
wavelength) and changes direction
(bends towards the normal)
Diffraction: Water passes through
a gap in a concrete bridge causing
a circular wave front. The longer
the wavelength (or smaller the
gap) the more it spreads out.
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12.4 Wave properties 2
Spec 3.2.3: Superposition of waves,
stationary waves; Interference
Be able to clearly explain the
concept and features of…
1.
How can we explain sound waves?
A03c – Link the idea of sound waves
in musical instruments to different
frequencies .
2.
What features of two waves must
combine in order to produce
reinforcement?
A01a –superposition, supercrests,
super troughs, cancellation. (Basic)
3.
What is the phase difference between
two waves if they produce maximum
cancellation?
A01a – Use a virtual ripple tank to
show interference patterns of two
circular waves. (Harder)
A03a – Apply ideas to a microwave
transmitter – (Basic)
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The Trumpet
Trumpet
Chromatic
Scale
Period ms
Bb
C
4
B
C#
277
C
D
293
C#
Eb
311
D
E
329
Eb
F
349
E
F#
F
G
392
F#
Ab
415
3
Frequency Hz Frequency Hz
(Calculated)
250
333
261
370
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Checking a Guitar’s Tuning.... Extension!
Guitar
Period ms
Frequency Hz
(Calculated)
Frequency Hz
E
0.00525
190
41
A
0.01
100
55
D
0.012
83
73
G
0.0125
80
98
String
Note
Frequency
1 (thinnest)
G3
97.999 Hz
2
D3
73.416 Hz
3
A2
55 Hz
4 (thickest)
E2
41.204 Hz
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Thick
thin
Index
The Real World - Extension
A tuning fork produces a note with only
one frequency. The shape of the wave on
the oscilloscope is very smooth.
However, the frequency of the harmonics
in a real instrument may be twice, three
times, four times or even more times the
fundamental frequency.
All these frequencies together make up
the note.
The bottom line here shows the wave
pattern formed by the fundamental and
harmonic frequencies when the note is
played on the instrument.
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Real Sounds - Extension
We now know that we can convert
our longitudinal sound wave to a
transverse wave to show on a
screen.
If we look at these three traces of
a middle C note (261Hz) we can
see they are all different but seem
to have similar pattern in terms of
frequency as.......
1 up and 1 down takes (1/261)th
of a second or the length of an
arrow!
You need to try an ignore the
funny fluctuations, this is due to
the timbre of the notes – or
richness that some from the
instrument itself
clarinet
violin
saxophone
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Definitions...
A progressive wave is one where the waveform travels, as
opposed to a standing wave (or stationary wave) where
the waveform is fixed in place.
Most familiar waves are usually progressive: light, sound,
and water transmit energy along their direction of travel,
though it is possible to set up standing waves for each of
these.
A plucked string fixed at both ends vibrates in a standing
wave though the musical sound it generates is a
progressive wave.
Progressive waves, despite the name, can travel
backwards as well as forwards. A standing wave is
equivalent to two equal and opposite progressive waves.
It can be either a transverse wave or a longitudinal wave,
depending on which direction the vibrations go compared
to the direction of travel of the wavefront. The wavefront
represents the pattern that is moving along.
TASK...
Use this
information to
explain where you
might find a
progressive wave
and how you can
create a standing
wave. Give an
example of each.
You can also refer
to your book as
well.
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Principal of Superposition
The resultant displacement at any point is the sum of the separate
displacements due to the two waves Eg: with a slinky coil spring
supercrest
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Two square waves superposing:
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Superposition of sine waves:
Fundamental
frequency
3*fo
A square wave can be made
up from several sine waves
of higher frequencies
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Phase Changes in Reflection
TRANSVERSE PULSE
LONGITUDINAL PULSE
NB: unlikely to be asked about this for Longitudinal (just Transverse)
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Reasons for behaviour of longitudinal – (Extension)
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Interference types....
Constructive
Destructive
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Interference – from previous lesson…
Two dippers in a ripple tank can cause circular
wavefronts to re-inforce or cancel.
Use the virtual ripple tank..
W:\Students Read Only\Physics\AS\Unit 12\ripple
tank\index.html
To explain the idea and draw a diagram to explain the
ideas in the purple boxes.
Re-inforcement
(constructive interference)
Cancellation
(destructive interference)
Coherent sources (of the same frequency and phase
relationship) produce a stable interference pattern.
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Regions of
reinforcement
Regions of
cancellation
Experiments with microwaves:
a) The intensity of the receiver signal decreases with distance from the transmitter.
b) Microwaves are reflected off metal plates – similar to light on a mirror.
c) Diffraction occurs at each slit (slit width is of similar magnitude to the
wavelength)
d) An interference pattern forms with regions of constructive and destructive
interference
Index
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Two loud speakers
emitting the same
note can cause
loud and quiet
areas in front
of the speakers
rarefaction
When compressions
(or rarefactions)
arrive in phase from
both speakers,
constructive
interference occurs,
creating a loud region
compressions
Regions of reinforcement (LOUD)
Regions of cancellation (QUIET)
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Plenary / Starter
1) Can you draw a diagram to show
how two waves meeting can…
a)
Destructively interfere?
b) Constructively interfere?
2) When exploring interference why
would you pass microwaves
through two slits?
3) What two conditions are required
for this pattern as shown to be
seen and be stable?
Diagram similar to show…
a)
+ = --
b)
+=
2) Create two sources of the same
frequency.
3) You need coherence i.e. same
frequency and phase difference
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12.5 Stationary and progressive waves
1.
What are the necessary conditions for the
formation of a stationary/standing wave?
2.
What part does superposition play?
3.
Why are nodes formed in fixed positions?
phase shift
2d
A03b – Take readings to work out the speed of sound from a
standing wave. Practical & Q1 (Basic)
A01a – Be able explain the idea of nodes (fixed), antinodes (max
amplitude) (Basic)
A01a – Explain that a stationary wave is formed from two
progressive waves in opposite directions. (Medium/Harder)
A01a/2b – Complete phase calculations for a stationary wave and
phase differences (Q2/Q3/Q4) (Medium /Harder)
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Practical Investigation...
1. Take your ruler and investigate the sound wave it
creates by “twanging” it with your fingers. (Take care
not to break it)
2. Think about the relationship between pitch (frequency)
and length.
3. Then make a verbal prediction for what might happen
with a string or tube?
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Why are nodes formed in fixed positions? “ Finding the Speed of
Sound”
Experiment…
You have the equipment shown in
the diagram and also a selection of
tuning forks and tube sizes.
Record your results in a suitable
table and see if you can work out the
speed of sound….
Ensure you quote any errors
(converting to an overall error) and
write a conclusion about your
results.
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Speed of Sound – Extra Help…
You might use a table as shown below…
Also you might decide to repeat your readings for
more accuracy.
Conclusions might talk about what happens to the
velocity of the various waves?
343 dry air 20oC
Frequency
(Hz)
Length (m)
Wavelength
(m)
Velocity
(ms-1)
Max amp at end
/4
Ave
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5/2
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Speed of Sound – Extra Help…
You might use a table as shown below…
Also you might decide to repeat your readings for
more accuracy.
Conclusions might talk about what happens to the
velocity of the various waves?
343 dry air 20oC
Max amp at end
Frequency
(Hz)
Length (m)
Wavelength
(m)
Velocity (ms-1)
256
0.34
1.36
348
271.2
0.300
1.200
325
304.4
0.285
1.140
335
320.0
0.250
1.000
320
341.3
0.229
0.916
313
Ave
324
/4
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5/2
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Data Trends... (Extension)
Discuss this data with a partner. Can
you see a trend in the numbers?
Can you comment on...
Gas -> Liquid -> Solid
the mass of the molecules or
compounds? (as best you know)
Ethanol
Chloroform
Glass
C2H5OH
CHCl3
SiO2
The bonding or strength of the
structures
You can use the periodic table to
help you?
Think helium and voice box (fixed )
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Wave examples Progressive or Standing....
Cornstarch is a shear thickening nonNewtonian fluid meaning that it
becomes more viscous when it is
disturbed. When it's hit repeatedly by
something like a speaker cone it forms
weird tendrils. The speaker cone was
vibrating at 30 Hz.
Jelly A large cubic shape shot
by a BB gun.
The Rubens Tube The classic physics
experiment involving sound, a tube of
propane and fire. Push the tube to
449 Hz then higher frequencies, then
some jazz and then some rock. This is
real life sound visualization....
Chladni plate: Fine sand
sprinkled on the plate gathers
at the nodes. Similar to a
wobble card (Rolf Harris)
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How Science Works – “Ernst Chladni”
Ernst Florens Friedrich Chladni (German pronunciation:
November 30, 1756 – April 3, 1827) was a German physicist
and musician.
His important works include research on vibrating plates
and the calculation of the speed of sound for different
gases.
One of Chladni's best-known achievements was inventing a
technique to show the various modes of vibration on a
mechanical surface.
Chladni repeated the pioneering experiments of Robert
Hooke of Oxford University who, on July 8, 1680, had
observed the nodal patterns associated with the vibrations
of glass plates.
Hooke ran a bow along the edge of a plate covered with
flour, and saw the nodal patterns emerge.
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Definition of a Standing Wave…
A vibration of a system in which some
particular points remain fixed while others
between them vibrate with the maximum
amplitude or less.
TASK: A01a – Use the
information shown to explain
the idea of nodes and
antinodes on a standing wave
in your own style. (Basic)
/4
3/2
5/2
The fixed points are called “nodes”
the moving maximum amplitudes
“antinodes”
Latin: nodus, 'knot'
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How is a stationary wave formed?
Hopefully you can see that as the wave passes through the reflected wave they
cancel at certain points only where the phase is matched.
This animation really shows it well as the blue/red waves interfere to produce the
black wave. Then you can see the nodes fixed where there is no movement...
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How is a stationary wave formed?
A progressive wave travels to a
wall (i.e. on a string) which is then
reflected back to the point of
origin.
out
phase
We then get a “standing wave”
formed. Where pattern is fixed.
The amplitude varies from
position from zero to +/- A
so some particles will not
move at all unlike a
progressive wave when all
move by +/-A
phase shift
2d
in phase
The phase difference between particles is….
1.
Zero between adjacent nodes, or an
even number of nodes.
2.
180 or radians if two particles are
separated by an odd number of nodes
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Is a stationary wave formed by superposition?
Yes it is and you can see by looking at this
graphic. The two waves interfere when
the meet.
1.
They can either constructively add
together
2.
Destructively cancel
3.
Work to some compromises.
Look the combined wave trace for each
case by passing a ruler through your
printed copy.
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How is a stationary wave formed?
A stationary wave is formed when….
The Amplitude….
TASK:
The Phase difference between particles is….
A01a – Explain how a
stationary wave is formed
from two progressive waves in
opposite directions.
(Medium/Harder)
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Extra Help…
1. Imagine you are holding a rope at one end which is attached to a brick wall at the other.
2. You are sending regular oscillations down the rope and something weird is happening.
You cannot see the top image but only the bottom one.
Extra words to help...
Frequency, amplitude, phase, node, antinode, super crest,
super trough, reflection, cancel, reinforce, destructive,
constructive, superposition, 180,
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Q3
In a stationary wave all the particles in a between a node are acting in phase..
i.e. they travel up at the same time. In a progressive wave each part of the
wave is out of phase as you move along the wave through 360…….
phase shift
2d
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Exam Question…
Explain the differences between an undamped progressive transverse wave and a
stationary transverse wave, in terms of (i) amplitude, (ii) phase and (iii) energy transfer.
(Total 5 marks)
Progressive Wave
Stationary Wave
Amplitude (2)
Phase (1)
Energy
Transfer (2)
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Exam Question Extra Help…
Explain the differences between an undamped progressive transverse wave and a
stationary transverse wave, in terms of (i) amplitude, (ii) phase and (iii) energy transfer.
(Total 5 marks)
Progressive Wave
Amplitude (2)
each point along wave
has same amplitude for
progressive wave
between nodes all particles
vibrate in phase
Phase (1)
Energy
Transfer (2)
Stationary Wave
energy is transferred through
space
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Exam Question…
Explain the differences between an undamped progressive transverse wave and a
stationary transverse wave, in terms of (i) amplitude, (ii) phase and (iii) energy transfer.
(Total 5 marks)
Progressive Wave
Stationary Wave
Amplitude (2)
each point along wave
has same amplitude
each point along wave
varies for stationary wave
Phase (1)
adjacent points vibrate with
different phase
between nodes all particles
vibrate in phase
Energy
Transfer (2)
energy is transferred through
space
energy is not transferred
through space (due to
opposing progressive waves)
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Exam Question – Starter/ Plenary
P
Q
R
S
1.20 m
The diagram represents a stationary wave
on a stretched string. The continuous line
shows the position of the string at a
particular instant when the displacement is
a maximum. P and S are the fixed ends of
the string. Q and R are the positions of the
nodes. The speed of waves on the string is
200 ms–1.
(i) State the wavelength of the waves on the
string. (1mark)
(ii) Calculate the frequency of vibration. (1
mark)
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Exam Question - Starter
P
Q
R
S
1.20 m
The diagram represents a stationary wave on a
stretched string. The continuous line shows the
position of the string at a particular instant when
the displacement is a maximum. P and S are the
fixed ends of the string. Q and R are the positions
of the nodes. The speed of waves on the string is
200 ms–1.
(i) State the wavelength of the waves on the string.
(1mark)
Answers
i) 0.8m (1)
ii)
Use of f = c/
f = 200ms-1/0.8m
= 250Hz
ecf.
(ii) Calculate the frequency of vibration. (1 mark)
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12.6 Stationary waves on strings
1.
What boundary condition must be
satisfied at both ends of the string?
2.
What is the simplest possible stationary
wave pattern that can be formed?
3.
How do the frequencies of the overtones
compare with the fundamental
frequency?
A02b – Compare theoretical table for overtones and complete the
table (Harder/Medium)
A03b – Take readings to work the fundamental frequencies of a string.
Demo / Table & Q1 (Medium) (Also apply f = mc/2L)
mc
f
2L
A02b – Exam Question for Guitar (Basic / Medium)
Mr Powell 2011
Index
Why are nodes formed in fixed positions?
• We are in effect looking at interference
points. Of which there must be a fixed point
at either end then a whole number in
between.
• The first or “fundamental pattern” can be
found as 0.
• Then it goes up by one antinode at a time for
each overtone.
0 2 L
f0
c
0
c
f0
2L
• You then simply sub in the fN for f in the usual
formula
f1 1 L
2L
f 2 2
3
2L L
f 3 3
4
2
Mr Powell 2011
Index
String Theory! (Not that one)
2L m
c
f
mc
f
2L
A wave on a string must take a time t
to travel along and return the
length of a string…
t = 2L/c
Where L is the length of the string
and c is the speed of the wave.
The time taken for the mechanical
oscillator to pass through a whole
number of cycles is….
t = m/f
So we can express
the situation to
find L in terms of
lambda and m.
“m” is the number and “f” is cycles per
second.
NB: “m” also represents the number of antinodes
Mr Powell 2011
mc
f
2L
2L c
m
f
2L
m
m
L
2
Index
Summary
Example….
Now we can use the formulas we have created
from first principals to conclude that due
to…
mc
f
2L
Stationary waves fitting these conditions (travel
back and forth along string) are formed in
multiple frequencies….
f0, 2f0, 3f0, 4f0
m = 1 fundamental frequency
m = 2 frequency 2 x f0
m = 3 frequency 3 x f0
Example….
m = 1 *1/2
m = 2 *2/2 =
m = 3 *3/2 = 1.5
Also that the length of the vibrating section of
the string is a whole number of half
wavelengths….
L
m
2
NB: “m” also represents the number of antinodes
Mr Powell 2011
Index
Working it all out…
Harder
Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
fo
AntiNodes
(m)
2L/m =
1
2L/1 =
2L
Freq
equation
f = c/
f0 = c/2L
Freq c = f Speed
value
(ms-1)
L = length of string
fundamental
256
410
1st overtone
2fo
2nd overtone
3fo
3rd overtone
4fo
4th overtone
5fo
Mr Powell 2011
Index
Working it all out…
Medium
Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
fo
AntiNodes
(m)
2L/m =
1
2L/1 =
2L
Freq
equation
f = c/
f0 = c/2L
Freq c = f Speed
value
(ms-1)
L = length of string
fundamental
256
410
1st overtone
2fo
2
f1 = c/L
410
2nd overtone
3fo
2L/3
f2 = 3c/2L
768
3rd overtone
4fo
f3 = 2c/L
4th overtone
5fo
Mr Powell 2011
Index
Working it all out…
Complete
Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
fo
AntiNodes
(m)
2L/m =
1
2L/1 =
2L
Freq
equation
f = c/
f0 = c/2L
Freq c = f Speed
value
(ms-1)
L = length of string
fundamental
256
410
1st overtone
2fo
2
2L/2 = L
f1 = c/L
512
410
2nd overtone
3fo
3
2L/3
f2 = 3c/2L
768
410
3rd overtone
4fo
4
2L/4 =
L/2
f3 = 2c/L
1025
410
4th overtone
5fo
5
2L/5
f4 = 5c/2L
1281
410
Mr Powell 2011
Index
Data Table?
NB: m is number of antinodes!
2L
m
Frequency
String Length
f0
Number of
anti-nodes
c
0
Lambda
Wave Speed
Ave
Mr Powell 2011
Index
Data Table?
NB: m is number of antinodes!
Frequency
String Length
Number of
nodes
Lambda
Wave Speed
8
2.62
2
2.62
21
16
2.62
3
1.75
28
24.3
2.62
4
1.31
32
32.5
2.62
5
1.05
34
38.5
2.62
6
0.87
34
Ave
30
2L
m
f0
c
0
Mr Powell 2011
Index
Example Results?
Frequency
String Length
Nodes
Lambda
Wave Speed
18
28
43.4
57.9
71.1
83.1
95.4
2.42
2.42
2.42
2.42
2.42
2.42
2.42
1
2
3
4
5
6
7
2.42
1.61
1.21
0.97
0.81
0.69
0.61
Ave
43.6
45.2
52.5
56.0
57.4
57.5
57.7
52.8
2L
m
f0
c
0
Mr Powell 2011
Index
b)
c = f
= 1.6m
f = 256Hz
a)
c = f
2L
m
c = f
= 256Hz x 1.6m
= 256s-1 x 1.6m
= 409.6ms-1
= 410ms-1 (3 sf)
l = 0.8m
f = 256Hz
The fundamental mode
must be node to node
so is only ½
fundamental
1st overtone
2nd overtone
Thus = (2 x 0.8m) / 1
= 1.6m (2 sf)
Mr Powell 2011
Index
b) c = 410ms-1 , f = 384Hz, m = 1
a) c = 410ms-1 , f = 512Hz, m = 1
(assume that this is a
fundamental so m = 1)
f = mc / 2L
L = mc/ 2f
L = 410ms-1 / (2 x 384Hz)
= 410ms-1 / (2 x 384s-1)
= 0.53385m
= 0.53m (2sf)
fundamental
f = mc / 2L
L = mc/ 2f
L = 1* 410ms-1 / (2 x 512Hz)
= 1* 410ms-1 / (2 x 512s-1)
= 0.40039m
= 0.40m (2sf)
1st overtone
2nd overtone
Mr Powell 2011
Index
Exam Question
The image shows a side view of a string on a guitar. The string
cannot move at either of the two bridges when it is vibrating.
When vibrating in its fundamental mode the frequency of the
sound produced is 108 Hz.
(a)
Sketch the stationary wave produced when the string is
vibrating in its fundamental mode. (1 mark)
(b) Calculate the frequency of the overtone. (1 mark)
c) Calculate the speed of a progressive wave on this string.
(1mark)
Mr Powell 2011
Index
Exam Question
The image shows a side view of a string on a guitar.
The string cannot move at either of the two bridges
when it is vibrating. When vibrating in its
fundamental mode the frequency of the sound
produced is 108 Hz.
(a) Sketch the stationary wave produced when the
string is vibrating in its fundamental mode. (1
mark)
(b) Calculate the frequency of the overtone. (1 mark)
c) Calculate the speed of a progressive wave on this
string. (1mark)
Answers
a) one ‘loop’ nodes at A &
B
b) f1= 2 x f0 = 2 x 108Hz
= 216Hz
c) λ0 = 2L or λ = 0.64 × 2
= 1.3m or 1.28m
c = f λ = 108Hz × 1.3m
= 138 to 140ms-1
ecf
Mr Powell 2011
Index
Revision/ Extension…
Visit the sites show below and then write a paragraph and numerical example
for each.
Inverse square law calculation:
http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/isprob.html#c3#
Geological example of sound reflection:
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/mamlak.html#c2
Pitch:
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/pitch.html#c1
Loudness of wave:
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/loud.html#c1
String properties:
http://hyperphysics.phy-astr.gsu.edu/hbase/music/stringa.html#c1
Wave properties and more :
http://www.glenbrook.k12.il.us/gbssci/phys/Class/sound/soundtoc.html
Mr Powell 2011
Index
Tension of Wire – Extension?
By increasing the frequency of the vibrator
Different stationary wave (s.w.) patterns
are seen.
Boundary condition for s.w. on a string is that
there must be a node at each end.
Velocity of a transverse wave in a wire or
string:
We find that....
c T
1
c
T = Tension (N)
c
T
= mass/ unit length kg/m
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/wavsol.html#c2
Mr Powell 2011
Index
More on Strings…
How the fundamental frequency of a vibrating string depends on the string's length, tension,
and mass per unit length is described by three laws:
1. The fundamental frequency of a vibrating string is inversely proportional to its length.
Reducing the length of a vibrating string by one-half will double its frequency, raising the
pitch by one octave, if the tension remains the same.
2. The fundamental frequency of a vibrating string is directly proportional to the square root
of the tension. Increasing the tension of a vibrating string raises the frequency; if the
tension is made four times as great, the frequency is doubled, and the pitch is raised by
one octave.
3. The fundamental frequency of a vibrating string is inversely proportional to the square
root of the mass per unit length.
This means that of two strings of the same material and with the same length and tension,
the thicker string has the lower fundamental frequency. If the mass per unit length of one
string is four times that of the other, the thicker string has a fundamental frequency one-half
that of the thinner string and produces a tone one octave lower.
Mr Powell 2011
Index
b) the frequency of a vibrating string is directly
proportional to the square root of the tension.
f T or
a) A shorter string or length
would yield a higher frequency
as the frequency is inversely
proportional to its length.
f = mc / 2L or f 1/L
If you half the length you will
double its frequency.
T f2
Thus if the tension is increased the frequency
(pitch) goes up as a square relation.
c f
f
c
c
2L
2 Lf 0 c
f0
T = Tension (N)
m = string mass kg
L = string length (m)
= mass/ unit length kgm-1
also...
This raises the pitch by one
octave on a musical scale.
Tension would be the same.
NB: This moves into
extension work!
c
T
m/ L
c
T
2 Lf 0
m/ L
f0
T
m/ L
2L
Mr Powell 2011
T
Index
T = Tension (N)
m = string mass kg
L = string length (m)
= mass/ unit length kgm-1
c
T
a) It would make sense that if L, T, diameter are fixed the steel wire being
more dense and would have a larger mass.
This would mean that as c2 1/m as the mass went up the wave would
travel slower through the wire.
Hence steel is slower than nylon (as it is more dense – makes sense).
As c = f as c goes down so does f as is fixed from the length of the wire.
So steel strings would be a lower pitch for same conditions as nylon.
Mr Powell 2011
c
T
c
T
m
L
TL
c
m
TL
2
c
m
1
2
c
m
Index
Further Reading on Tension & Frequency….
Hanging Mass
3 kg
4 kg
5 kg
6 kg
Tension
29.4 N
39.2 N
49 N
58.8 N
Frequency
98 Hz
112 Hz
122.5 Hz
131 Hz
A brief data set with a steel string is
plotted to check the consistency of
the data with the equation shown.
Hanging masses provide the tension
in the string, which was adjusted to
a vibrating length of 50 cm. The
average slope of the line through
zero is taken, the slope is 17.6 HzN0.5.
Using the frequency relationship
above, this corresponds to a mass
per unit length m/L = 57
grams/meter.
Source: Hyper Physics
c
T
Key Summary
1. the shorter the string, the higher the
frequency of the fundamental
2. the higher the tension, the higher the
frequency of the fundamental
3. the lighter the string, the higher the
frequency of the fundamental
Mr Powell 2011
Index
Extension Question
A wire of mass per unit length 0.5 x 10-3 kgm-1 has a tension of 60 N and is
50 cm long.
i.
ii.
iii.
iv.
Calculate the velocity of any transverse wave in the wire
Calculate the frequency of the fundamental note.
If nodes are at 17cm and 34cm find the frequency of the vibrating
wire.
What must the tension be if a note an octave above the fundamental is
required? ( 2x fundamental frequency)
Mr Powell 2011
Index