Standing waves in an AC

1. Lesson #6Lesson #6
Standing Waves in an Air Column
Nelson Reference Pages:457-458Nelson Reference Pages:457-458
(Nelson does not cover all topics)(Nelson does not cover all topics)
McGraw-Hill Reference Pages: 390-398McGraw-Hill Reference Pages: 390-398

2. Forming a Standing Wave in an Air
Column Closed at One End
 If you have ever blown into aIf you have ever blown into a
pop bottle and heard apop bottle and heard a
humming sound, then you havehumming sound, then you have
created a standing wave in ancreated a standing wave in an
AC. The following diagramAC. The following diagram
shows how the wave formedshows how the wave formed
can be represented.can be represented. Antinodes
represent high and lowrepresent high and low
pressure (compression &pressure (compression &
rarefaction) andrarefaction) and nodes
represent normal (& constant)represent normal (& constant)
pressure.pressure.
L
A ir C o lu m n C lo s e d
a t O n e E n d
A n t in o d e
N o d e

3.  If we try to relate theIf we try to relate the
length of the AC tolength of the AC to λλ, we, we
would getwould get λλ = __ L. This= __ L. This
is lowest naturalis lowest natural
frequency that canfrequency that can
occur. It is calledoccur. It is called
_________ frequency or_________ frequency or
________ harmonic.________ harmonic.
 We can see why this isWe can see why this is
true from the diagram.true from the diagram.
 Note, the wave does notNote, the wave does not
extend out of the AC asextend out of the AC as
shown in the diagram.shown in the diagram.
L
A c o m p le t e
w a v e is s h o w n

4. Other Resonant Lengths for the Same
Frequency
 We will modify the previous drawingWe will modify the previous drawing
to show how we can determine theto show how we can determine the
next resonant length for thenext resonant length for the same
frequency.frequency.
 We know that there must be anWe know that there must be an
antinode at the opening whichantinode at the opening which
required addingrequired adding ½ λ
 The equation for other resonantThe equation for other resonant
lengths (for same f andlengths (for same f and λλ) would be:) would be:
Ln = (2n -1) ¼λ, n = 1,2,…
Here, when n = 1 we have the shortestHere, when n = 1 we have the shortest
length and this corresponds to thelength and this corresponds to the
fundamental frequency (firstfundamental frequency (first
harmonic).harmonic).
L
A c o m p le t e
w a v e is s h o w n
3 L

5. Other Natural Frequencies for a
Given Length of AC Closed @ 1 End
 Not all harmonics occur in a closed AC, onlyNot all harmonics occur in a closed AC, only
the odd numbered ones; 1,3,5.. This is whythe odd numbered ones; 1,3,5.. This is why
the Fthe Fnn equation at the bottom has “2n-1”equation at the bottom has “2n-1”
which only gives an odd number when “n” iswhich only gives an odd number when “n” is
a positive integer.a positive integer.
 Purple represents the fundamental frepresents the fundamental f11. We. We
know that v=fknow that v=f λλ, so f, so f11 =v /=v / λλ, but, but λλ = 4L, so f= 4L, so f11
= v/(4L)= v/(4L)
 Green represents therepresents the second overtoneovertone (or 3rd
harmonic). We can see that L = ¾We can see that L = ¾ λλ, so, so λλ
= (4/3) L and f = v /((4/3)L) = ¾ v/L= (4/3) L and f = v /((4/3)L) = ¾ v/L
 Blue represents therepresents the 4th
OTOT (or 5th
harmonic).
 The general equation is:The general equation is:
Fn = (2n-1) x f1 , n = 1,2,… think of “n” as
representing the next lowest natural
frequency that will occur (it is NOT the
harmonic number – which equals “2n-1” ). L

6. Example Problem
 A train whistle has a length of 0.600 m andA train whistle has a length of 0.600 m and
is closed at one end. If air temp. is 15is closed at one end. If air temp. is 1500
C,C,
determine the three lowest resonantdetermine the three lowest resonant
frequencies.frequencies.
 Ans: ff1 = 142 Hz,= 142 Hz, ff2 = 425 Hz,= 425 Hz, ff3 = 708 Hz= 708 Hz
 Here, f2 would be the 2nd
lowest natural
frequency but represents the 3rd
harmonic,
similarly f3 would be the 3rd
lowest natural
frequency but represent the 5th
harmonic.
(Note: 2nd
and 4th
harmonics can not form in
the closed AC)

7. AC Opened at Both Ends
 From the diagram, we can see thatFrom the diagram, we can see that
an antinode (an antinode (A) must occur at each) must occur at each
end.end.
 For the fundamental frequency (fFor the fundamental frequency (f11),), λλ
= __ L. We can see that for the next= __ L. We can see that for the next
resonant length (Lresonant length (L22), __), __ λλ must bemust be
added to the initial length (Ladded to the initial length (L11) of the) of the
AC. The equation for the resonantAC. The equation for the resonant
lengths is:lengths is:
Ln = n x ½ λ, n = 1, 2,…
(Where L1 is the shortest length and
results in the fundamental frequency
– see diagram)
A C o p e n e d a t
b o t h e n d s .L
A
N
A

8. Three Lowest Natural Frequencies for
a Given Length of Open AC
 Again we use theAgain we use the
fundamental wave equation,fundamental wave equation,
wherewhere v is a constant for ais a constant for a
given temperature. For thegiven temperature. For the
first OT,first OT, λ is now half ofis now half of
what it was forwhat it was for f1 , this means, this means
that fthat f22 must bemust be 2 x f1..
SimilarlySimilarly F3 = 3f1..
 The equation for naturalThe equation for natural
frequencies for a givenfrequencies for a given
length is:length is:
 Fn = n x f1 , n = 1,2,..
L F u n d a m e n t a l
 1 s t O v e r t o n e
 2 n d O v e r t o n e

9. Examples of AC’s
 Examples of Open AC’s: At the CNE,Examples of Open AC’s: At the CNE,
some prizes have been plastic tubes thatsome prizes have been plastic tubes that
you whirl above your head. These areyou whirl above your head. These are
open at both ends. A bugle flute is also anopen at both ends. A bugle flute is also an
open AC.open AC.
 Reed instruments such as clarinet, oboe,Reed instruments such as clarinet, oboe,
bassoon, behave like an AC closed at onebassoon, behave like an AC closed at one
end.end.

10. Practice Problems
The following website explains how pressure wavesThe following website explains how pressure waves
(compression & rarefaction) occur in a closed AC. It also(compression & rarefaction) occur in a closed AC. It also
explains why only odd numbered harmonics occur (1,3,5…)explains why only odd numbered harmonics occur (1,3,5…)
http://www.webtech.buffalo.edu/student/m3_digimusic/02_documents/sound_wave/MusicInstruments_ClosedEn
Nelson TB: Page 460 #7, 8, 9, 10Page 460 #7, 8, 9, 10
Questions from the McGraw-Hill TB (Page 396)
1. A hollow plastic pipe is almost completely submerged in a1. A hollow plastic pipe is almost completely submerged in a
graduated cylinder full of water, and a vibrating tuning forkgraduated cylinder full of water, and a vibrating tuning fork
is held over its open end. The pipe is slowly raised from theis held over its open end. The pipe is slowly raised from the
water. An increase in loudness of the sound is heard whenwater. An increase in loudness of the sound is heard when
the pipe has been raisedthe pipe has been raised 17 cm and then raised byand then raised by 51 cm..
a.a. What does an increase in loudness represent?What does an increase in loudness represent? (Ask)
b.b. Determine the wavelength of the sound produced.Determine the wavelength of the sound produced. (68cm)
c.c. If the pipe continues to be raised, how far is the top of theIf the pipe continues to be raised, how far is the top of the
pipe from the water when the next increase in loudness ispipe from the water when the next increase in loudness is
heard?heard? (85cm)

11. 2. The first resonant length of an air column
resonating to a fixed frequency is 32 cm.
a. Determine L2 and L3 if the AC is closed at
one end. (96cm, 160 cm)
b. Same as part “a” except column is open at
both ends. (64cm, 96cm)
3. The third resonant length of a closed air
column resonating to a tuning fork is 95
cm. Determine the first and second
resonant lengths. (19 cm, 57cm)

12. 4. A pipe organ, open at both ends, must resonate at 128 Hz.
The organ has been designed to be played at a
temperature of 22 0
C.
a. How long does the organ pipe need to be? (Assumption?)
b. If one end is closed by a stopper, at what new
fundamental frequency will it resonate? (1.34 m, 64 Hz)
5. An AC open at both ends has a fundamental frequency of
256 Hz. Determine the frequencies of the first and
second overtones. (512 Hz, 768 Hz)
6. A slightly smaller plastic pipe is inserted into a second
plastic pipe and the resulting variable length AC is open
at both ends. A tuning fork of unknown frequency is
allowed to vibrate over one end of the AC and resonance
is heard when the AC is 38 cm long and then again when
it is 57 cm long.
a. Calculate the wavelength of the vibration created by the
tuning fork. (38 cm)
b. If the air temperature is 180
C, determine the frequency of
the tuning fork. (9.0x102
Hz)