PHYS1002
Physics 1
FUNDAMENTALS
Module 3
OSCILLATIONS & WAVES
Text
Physics by Hecht
Chapter 11
WAVES
Standing Waves
Sections:
Doppler Effect
11.10 11.11
Examples: 11.12 11.13 11.14 11.15
CHECKLIST

Standing Waves (stationary waves)
interference, nodes, antinodes, wavelength  is twice the node-to-node
distance

Standing Waves on Strings - string fixed at both end
fundamental, harmonics, overtones, modes of vibration (Fig. 11.45)
Node Antinode N A N

L  N 
2
f N  N f1
v f 
FT
u
f1 
mode number, N = 1, 2, 3, …

1
2L
FT
u
  m/ L
(11.14, 11.15)
Standing Waves in air columns - both ends open or closed
fundamental, harmonics, overtones, modes of modal patterns, chamber open at
both ends and closed at both ends
pressure: open N A N A N open
closed
A N A N A closed
particle displacement: open A N A N A open

L  N 
2
f N  N f1
v f
closed
N A N A N closed
(11.14)
mode number, N = 1, 2, 3, …
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
Standing Waves in air columns - one end open, other closed
fundamental, harmonics, overtones, modes of vibration, modal patterns,
chamber open / closed
pressure: open N A N A closed
particle displacement: open A N A N A closed

L  N 
4
f N  N f1
v f
(11.16)
mode number, N = 1, 3, 5, … only odd harmonics resonant

Doppler Effect
fo  f s
v  vo
v  vs
(11.22)
fo > fs source and/or observing moving towards each other only
fo < fs source and/or observer receding from each other only
Shock waves
NOTES
STANDING (STATIONARY) WAVES

Wave travelling to right --->
y1(x,t) = A sin(k x -  t)

Wave travelling to the left <---
y2(x,t) = A sin(k x +  t)

Two waves travelling in opposite directions with equal displacement amplitudes
and with identical periods and wavelengths interfere with each other to give a
standing (stationary) wave (not a travelling wave - positions of nodes and
antinodes are fixed with time)

standing wave (no proof)
yR(x,t) = 2 A sin(k x) cos( t)
amplitude of standing wave (varies with position only and is independent of time)
yRm(x) = 2 A sin(k x) 
cos( t)  each point oscillates with SHM, period T = 2 / 
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
Nodes
If k x = n  (n = 0, 1, 2, 3, ...) then
yR(x) = 0  zero amplitude at x and at all times t
k = 2 /  (2 x) /  = n   x = n  / 2
 adjacent nodes are separated by  / 2

Antinodes
If k x = (n + ½)  (n = 0, 1, 2, 3, ...) then yR(x) = 2A
 max amplitude at x and at all times t
k = 2 /  (2 x) /  = (n + ½)   x = (n + ½) / 2
 adjacent antinodes are separated by  / 2 and are located halfway between
pairs of nodes
STANDING WAVES IN STRINGS AND RESONANCE
String instruments form a large group of musical instruments which include the
violin, guitar and piano. All these instruments make a sound by causing s taut string to
vibrate. the string may be bowed (violin), plucked (guitar) or struck by a hammer
(piano). The pitch of the note produced depends on three factors – length, linear
density and string tension. A shorter, lighter or tighter string gives a higher note.
The violin family of instruments are the most expressive of sting instruments. The
violin has four strings of different linear densities. These are wound around tuning
pegs to produce the correct tension. The performer stops the string vibrating to obtain
other notes. By pressing one or more strings against the fingerboard to shorten the
section that vibrates, higher notes can be played.
The body of the violin acts as an resonant amplifier. The front and back of the violin
are connected by a short sound post that transmits vibrations to the back. The whole
body vibrates and the sound wave is emitted through f-shaped sound holes on the
front of the instrument.
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Waves reflected at boundaries
O------------------------------------O
x=0
x=L

Boundary conditions imposed on vibrating string yR(0,t) = 0 and yR(L,t) = 0

The natural vibrations of the string are described by
yR(x,t) = 2 A sin(k x) cos(t)
At x = 0
yR(0,t) = 0  boundary condition satisfied
At x = L
yR(L,t) = 0  boundary condition  k L = N 
N = 1, 2, 3, ...
N is referred to as the mode number
The wavelength  is determined by the distance L between the supports at the end
2L
 
N 
L N N 
of the string
N
 2 
The speed of a transverse wave on a string is determine by the string tension and
FT
its linear density v 

fN 
Natural frequencies
 1  FT
 N

N
 2L  
v
Resonance (“large” amplitude oscillations) occurs when the string is excited or
driven at one of its natural frequencies.
 Harmonic series
N = 1 fundamental or first harmonic
N = 2 2nd harmonic (1st overtone)
N = 3 3rd harmonic (2nd overtone)
N
Nth harmonic (N-1 th overtone)
1 = 2L f1 = (1/2L).(FT / )
 2 = L = 1 / 2
f2 = 2 f1
3 = 2L / 3 = 1 / 3
f3 = 3 f1
N = 2L / N = 1 / N fN = N f1
N
10
120
100
4
80
3
60
40
2
20
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
position along string
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Why does a violin sound different to a viola?
Why do musicians have to tune their string instruments before a concert?
different string - 
Fingerboard
bridges - change L
tuning knobs (pegs)
- adjust FT
Body of instrument (belly)
resonant chamber - amplifier
L
f1 
1
2L
v
FT

FT

f 
f N  N f1
1
L
FT

N  1, 2,3,...
violin – spectrum
10
50
9
40
8
30
7
20
6
10
5
0
4
-10
-20
3
-30
2
-40
1
-50
0
1
2
3
4
5
6
7
8
9
0
10 11 12 13 14 15
0.002
0.004
0.006
0.008
0.006
0.008
time t (s)
harmonics (fundamental f 1 = 440 Hz)
viola – spectrum
14
50
40
12
30
10
20
10
8
0
-10
6
-20
4
-30
-40
2
-50
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
harmonics (fundamental f 1 = 440 Hz)
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0
0.002
0.004
time t (s)
5
Problem
A particular violin string plays at a frequency of 440 Hz. If the tension is increased
by 8.0%, what is the new frequency?
Solution
f1 = 440 Hz
f2 = ? Hz
FT2 = 1.08 FT1
1 = 2 1 = 2 L1 = L2
natural frequencies
N1 = N2
fN = N v / 2L = (N / 2L).(FT / )
f2 / f1 = (FT2 / FT1)
f2 = (440)(1.08) = 457 Hz
Problem
A guitar string is 900 mm long and has a mass of 3.6 g. The distance from the bridge
to the support post is 600 mm and the string is under a tension of 520 N.
1
2
3
4
Sketch the shape of the wave for the fundamental mode of vibration
Calculate the frequency of the fundamental.
Sketch the shape of the string for the sixth harmonic and calculate its
frequency.
Sketch the shape of the string for the third overtone and calculate its
frequency.
5
Solution
L1 = 900 mm = 0.900 m m = 3.6 g = 3.610-3 kg
L = 600 mm = 0.600 m
FT = 520 N
 = m / L1 = 3.610-3 / 0.9 = 0.004 kg.m-1
v = (FT / ) = (520 / 0.004) = 360.6 m.s-1
 1 = 2L = (2)(0.600) = 1.200 m
Fundamental frequency f1 = v / 1 = 360.6 / 1.2 = 300 Hz
fN = N f1
sixth harmonic N = 6
f6 = (6)(300) = 1800 Hz = 1.8 kHz
third overtone = 4th harmonic
N = 4
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f4 = (4)(300) = 1200 = 1.2 kHz
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STANDING WAVES IN AIR COLUMNS
Woodwind instruments are not necessarily made of wood eg saxophone, but they do
require wind to make a sound. They basically consist of a tube with a series of holes.
Air is blow into the top of the tube, either across a hole or past a flexible reed. This
makes the air inside the tube vibrate and give out a note. The pitch of the note
depends upon the length of the tube. A shorter tube produces a higher note, and so
holes are covered. Blowing harder makes a louder sound. To produce deep notes
woodwind instruments have to be quite long and therefore the tube is curved. Brass
instruments (usually made of brass) consist of a long pipe that is usually coiled and
has no holes. The player blows into a mouthpiece at one end of the pipe, the vibration
of the lips setting the air column vibrating throughout the pipe. The trombone has a
section of pipe called a slide that can be moved in and out. To produce a lower note
the slide is moved out. The trumpet has three pistons that are pushed down to open
extra sections of tubing. Up to six different notes are obtained by using combinations
of the three pistons.
Pipe closed at in end and open at the
other
closed end
particle displacement zero
 node
open end
max particle displacement
 antinode
120
100
80
60
40
20
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
position along column
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Boundary conditions - reflection of sound wave at end of air column (pipe)
At an open end – a compression is reflected as a rarefaction and a rarefaction as a
compression ( phase shift).
To match the boundary conditions
4L


2 N 1 
L   2 N  1  2 N 1 
N  1, 2,3,...
2N 1
 4 
Speed of sound in air (at room temperature v ~ 344 m.s-1) v = f 
Natural frequencies of vibration
f 2 N 1 
v
2 N 1
 2N 1 

v
 4L 
odd harmonics exit: f1, f3, f5, f7 , …
equilibrium position of particles
14
13
instantaneous position of
particles
12
11
10
sine curve showing
instantaneous displacement of
particles from equilibrium
9
8
7
6
instantaneous pressure
distribution
5
4
3
2
time averaged pressure
fluctuations
1
0
Enter t/T
-1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
For pipes closed at both end or open at both ends- all harmonics exists just like a
string fixed at both ends.

An air stream produced by mouth by blowing the instruments interacts with the air
in the pipe to maintain a steady oscillation.

All brass instruments are closed at one end by the mouth of the player.

Flute and piccolo – open at atmosphere and mouth piece (embouchure) – covering
holes L      f 

Trumpet – open at atmosphere and closed at mouth – covering holes adds loos of
tubing into air stream L      f 

Woodwinds – vibrating reed used to produce oscillation of the air molecules in the
pipe.
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Problem
What are the natural frequencies of a human ear?
Why do sounds ~ 3000 – 4000 Hz appear loudest?
Solution
Assume the ear acts as pipe open at the atmosphere and closed at the eardrum. The
length of the auditory canal is about 25 mm.
Take the speed of sound in air as 340 m.s-1.
L = 25 mm = 0.025 m
v = 340 m.s-1
For air column closed at one end and open at the other
L = 1 / 4  1 = 4 L  f1 = v / 1 = (340)/{(4)(0.025)} = 3400 Hz
When the ear is excited at a natural frequency of vibration  large amplitude
oscillations (resonance)  sounds will appear loudest ~ 3000 – 4000 Hz.
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DOPPLER EFFECT - motion related frequency changes
Doppler 1842, Buys Ballot 1845 - trumpeters on railway carriage
observer
v  vo
f o  fs
v  vs
v (source) = 0
v (source) = v (wave)
v (source) = v (wave) / 2
v (source) = 1.25 v (wave)
source vs
stationary
stationary
stationary
receding
approaching
receding
approaching
approaching
receding
observer vo
stationary
receding
approaching
stationary
stationary
receding
approaching
receding
approaching
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observed frequency fo
= fo
< fo
> fo
< fo
> fo
< fo
> fo
?
?
10
Applications: police microwave speed units, speed of a tennis ball, speed of blood
flowing through an artery, heart beat of a developing fetous, burglar alarms, sonar –
ships & submarines to detect submerged objects, detecting distance planets, observing
the motion of oscillating stars.
Shock Waves – supersonic waves
fast moving power boat
boat moving through water: speed of boat > speed of
water wave created
sailing boat rocket “violently” wave crests add to give “large wave”
shock wave - bunching of wavefronts ---> abruptive rise and fall of air pressure
plane travelling at supersonic speeds
Mach cone
Mach Number = v / vs vs speed of sound in air
eg Mach Number = 2.3 speed is 2.3 times the speed of sound
sonic boooom
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Problem
The speed of blood in the aorta is normally about 0.3000 m.s-1. What beat frequency
would you expect if 4.000 MHz ultrasound waves were directed along the blood flow
and reflected from the end of red blood cells? Assume that the sound waves travel
through the blood with a velocity of 1540 m.s-1.
Solution
Setup
fs1 = 4.0 MHz = 4.0x106 Hz
v01= 0.30 m.s-1
v = 1.54x103 m.s-1
fo1 = ? Hz
fo2 = ? Hz
fs2 = fo1
vs2 = 0.30 m.s-1
reflected wave
f o  fs
Doppler Effect
v  vo
v  vs
f beat  f 2  f1
Beats
Action
Blood is moving away from source  observer moving away from source  fo < fs
f o1  fs1
v  vo1
 4 106
v  vs1



10  0.30
  3.999221 10
  1.541.54
10
3
3


6
Hz
Wave reflected off red blood cells  source moving away from observer  fo < fs
f o2  fs2
v  vo2
 3.999221106
v  vs2


 1.54 103

6

  3.998442 10 Hz
3
 1.54 10  0.30 
Beat frequency = | 4.00 – 3.998442| 106 Hz = 1558 Hz
In this type of calculation you must keep extra significant figures.
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