IB PHYSICS NOTES and
STUDY GUIDE
TOPIC 4 Part 3
OSCILLATIONS and WAVES
NOTES
HL - SL
Standing Waves
1
1. Standing Waves FIXED ENDS pp. 251- 252
Standing waves are the result of the superposition of two waves
travelling in opposite directions.
A Standing wave is a wave pattern that does not appear to move in either
direction. Standing waves have variable amplitude and there
is no net transfer of energy in either direction. Travelling
wave – net transfer of energy(e.g ocean wave)
When a wave is reflected from fixed ends the incident and
reflected waves combine to form a stationary wave containing
nodes and antinodes as shown in Figure 11-39 (Giancoli
2005; 310). Nodes are points of destructive interference
that never move (the amplitude and thus, the energy in the
wave at that point is zero). Antinodes are points of
constructive interference where the string oscillates with
maximum amplitude and maximum energy. Nodes (and
antinodes) are equally spaced at ½ λ intervals.
 2 3
x  0, ,
, ,... (Location of nodes on a standing wave
2 2 2
with fixed end at x = 0)
Example 1 :
Which statement is true for standing (stationary) waves?
A.
All points in the wave vibrate in phase.
B.
There is no energy in a standing wave.
C.
The wavelength of the wave is the distance between adjacent nodes.
D.
Neighbouring points in the wave have different amplitudes of vibration.
Example 2 :
For a standing wave, all the particles between two successive nodes have the same
A.
amplitude only.
B.
frequency only.
C.
amplitude and frequency.
D.
frequency and energy.
2
String with Length L
Fixed at Both Ends (e.g. guitar, piano
string)
The frequencies at which standing waves are produced on a string are called the
natural frequencies or resonant frequencies of the cord. When a string is
plucked, waves of a great variety of frequencies will travel in both directions
along a string. Most of these waves interfere with each other and quickly die out.
However, those corresponding to the resonant frequencies of the string will
persist .The standing wave produced leads to a sound in air with a frequency
determined by the properties of the string.
fig. 6.2 tsokos p. 252
n = 1 ( 1 antinode) f0 (initial) 1st harmonic
The standing wave above has a specific wavelength (½ λ). Note that we have
fitted half a wave on the length of the string( L) . This means that :
L=λ
2
or
λ = 2L
note number of nodes = 2 and antinodes = 1
3
fig. 6.3 tsokos p. 252
n = 2 ( 2 antinodes) f2
2 nd harmonic
Fig. 6.3 shows the next standing wave. Note that the only constraint is that the
end of the string are nodes. Here we have fitted one full wavelength. This
means that :
L=λ
note number of nodes = 3 and antinodes = 2
An infinity of STANDING WAVES can thus exist on the string. The next standing
wave is shown below in fig. 6.4
Determine the formula for L in terms of λ and vice versa.
Determine the number of nodes and antinodes.
fig. 6.4 tsokos p. 252 n = 3 ( 3 antinodes)
f3
3rd
harmonic
4
Example : For n = 6 :
a) draw the standing wave ,
b) determine the formula for L in terms of λ and vice versa
c) determine the number of nodes and antinodes
Tsokos p. 256 : 1-4
5
L , λ , antinodes ( n)
A general formula can be used relating the length of the string (L), the
wavelength (λ) and the number of antinodes (n)
L
n
2
Where: n is a positive integer equal to the number of antinodes
n 
2L
n
Where: n = 1, 2, 3,…
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Frequency ( f) , harmonics and standing waves
Since f is inversely proportional to wavelength we will relate frequency to
standing waves using the terms harmonic and overtones. Harmonics , also
called overtones, are sounds based on a certain pitch or frequency.
Fundamental frequency ( f0 ) = 1st harmonic = 1st standing wave
n = 1 (1antinode)
L= nλ = ½λ
f = f0
2
If f0 is the fundamental frequency, then all other harmonics have frequencies
equal to integral multiples of f0.
f2 = 2nd harmonic = 2 f0
= 1st overtone
f3 = 3rd harmonic = 3f0 = 2nd overtone
f4 = ___________ = ___________ = ___________
General formula for 1st, 2nd, 3rd, harmonics….etc :
fn = n f0
7
f2 = 2nd harmonic = 2nd standing wave
n = 2 (2 antinodes )
L = n λ = 2λ 
2
L=λ
f2 = 2f0
2
f3 = 3rd harmonic = 3rd standing wave
n = 3 (3 antinodes ) L = n λ = 3λ
2
f3 = 3f0
2
Remember for constant velocity: as f increases λ decreases.
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Example : a) If a violin vibrates at 440Hz as its fundamental frequency determine
the first 4 harmonics.
b) If the length of string is 0.500m determine the wavelength of each harmonic.
v , λ , f , L and formula review
Since v = λ f
and λ = 2L
n
f = nv
2L
Also remember these :
fn = nf0
L = nλ
2
Length of string (L) , frequency ( f) and type of sound :




As L increases f decreases  low frequency bass sound (e.g. Cello).
As L decreases f increases high frequency treble sound (e.g. violin).
Guitar changes L by using the fret board.
Tuning changes the tension to correct the frequency.
9
Tsokos p. 256 : 10
Read Tsokos p. 254 : Q1, Q2 , p. 256 : 7 -9 f is proportional to √T (
tension )
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2. 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.
n 
2, 3,
fn 
2, 3,
2L
(Harmonic wavelengths, tube open at both ends, for any integer n = 1,
n
nv
(Harmonic frequencies, tube open at both ends, for any integer n = 1,
2L
Example :
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 ?
fn 
=
nv
2L
343
2 (0.26)
= 660Hz
2nd Harmonic = 2f0 = 1320Hz
3rd Harmonic = 3f0 = 1980 Hz
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Example : A pipe, open at both ends, 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
A.
v
.
2L
B.
L
.
2v
C.
4v
.
L
D.
L
.
4v
12
Tsokos p. 256 : 14 , 20 a
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3. 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 
3, 5,
4L
(Harmonic wavelengths, closed at one end, for any odd integer n = 1,
n
L = nλ
4
fn 
3, 5,
L = nv
4f
nv
(Harmonic frequencies, closed at one end, for any odd integer n = 1,
4L
Example :
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 
f0 =
nv
4L
343
4 (0.26)
= 330Hz
2nd Harmonic = 0 only odd harmonics exist
3rd Harmonic = 3f0 = 990 Hz
14
Tsokos p. 256 : 11
Resonance occurs at Antinodes:
Resonance and the Length of a Pipe or Tube Closed at One End
L = nv
L = nλ
4f
4
As mentioned 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, …).
Many questions ask where resonance will occur in terms of the length ( L) of the
tube.
Example: A tuning fork of 250 Hz is sounded at the top of a tube closed at one
end. If the tube has a length of 1.5 m where will resonance occur? Take speed of
sound = 330ms-1
Find L of the fundamental frequency ( 1st harmonic) f0 n = 1. Resonance will
occur at odd multiples of this length.
L = nv
4f
= 1( 330)
4(250)
= .33 m
So resonance will occur at odd multiples of 0. 33m : 0.33m, 0.99m, NOT 1.32mm
because 4th harmonic ( only odd harmonics produce resonance)
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Tsokos p. 256 : 12 , IB example
IB Example: A glass tube with one end open and the other closed is used in a resonance
experiment to determine the speed of sound. A tuning fork of frequency 427 Hz is
used and resonance is observed for an air column length equal to 17.4 cm. What
speed of sound does this experiment give?
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Closed at One End Tubes







vs.
Open Ended Tubes
An open ended tube can support any harmonic.
A closed end tube can only support odd harmonics.
A string with fixed ends and open ended tubes must have nodes at both
ends
closed tubes have a node at one end and an antinode at the other.
The distance between successive nodes is half a wavelength for
string/open tube
The distance between a node and an adjacent antinode is a quarter of a
wavelength for closed tube)
Sketching the first three standing waves should allow you to determine
the wavelength pattern relationship with length:
String/open tube :
Closed tube:
f open
vs.
L= n λ
2
L=nλ
4
n = 1, 2, 3, …
n = 1, 3, 5, ….
f closed
f open is proportional to
1/ 2L
f closed is proportional to 1/ 4L
f open = 2 f closed see example p. 22 
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Example:
Two pipes P and Q are of the same length. Pipe P is closed at one end and pipe Q is open at
both ends. The fundamental frequency (first harmonic) of the closed pipe P is 220
Hz.
The best estimate for the fundamental frequency of the open pipe Q is
A.
880 Hz.
C. 110 Hz
B.
440 Hz.
D. 55Hz
Additional Problems Below : 25, 30 a , b only, 33 a only pipe open at both
ends, 34 USE v = 330 ms-1
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