Sound Interference ----------------------------------------- Joseph F. Alward, PhD     Department of  Physics   University of the Pacific

Applet in this electure: Animated Standing Waves Superposition Superposition.

 Applets

Superposition Very good Superposition. Pulses, excellent

 

If speaker diaphragms are in phase, i.e., if they expand and contract at the same time, and the observer is equidistant from each speaker, the sound waves will be in phase

The two waves at the observer are always out of phase by one-half wavelength.   Sources are  coherent if the phase difference is always the same.

   Speakers Out of Phase

Each speaker gets two wires. If the connections don't match, speakers will be 180 degrees out of phase. A listener equidistant from each speaker would hear nothing.

f = 200 Hz         v = 343 m/s l = v / f = 343/200 = 1.7 m

d - x  =   l                         (1) d       =  x + l                   (2) d2  = 32 + x2                     (3) (x + l)2  = 9 + x2             (4) 2 x l + l2  = 9                   (5) x = (9 - l2) / 2l                  (6)    = (9 - 1.72) / 2(1.7)          (7)    = 1.8 m ------------------------------------ Condition for destructive inteference:  d - x = l/2

Electronic circuitry generates a rarefaction whenever a noise condensation is detected by the microphone, and a condensation whenever a rarefaction is detected.  The incoming noisy sound is canceled by the sound generated by the speaker inside the earphones.

 Introduction to Standing
 Sound Waves  


  Applet

Animated Standing Waves

 
Review of Sound Wave Condensations and Rarefactions

   Pressure Nodes and Antinodes

Pressure nodes and antinodes are separated by one-quarter wavelength, l / 4.

A pressure node is a displacement antinode. A pressure anti-node is a displacement node.

 Displacement Nodes and AntiNodes

Graph shows pressure, but labels inside tube refer to displacement nodes and antinodes. ------------------------------------------------ Pressure nodes are displacement antinodes, and pressure antinodes are displacement nodes. ------------------------------------------------ Displacement nodes and antinodes are separated by one-quarter wavelength, l / 4. ------------------------------------------------

Key Idea:  Nodes and antinodes are separated by one-quarter of a wavelength.

   Calculating Resonant Frequency

Problem: Sound waves are resonating in a standing wave pattern in the closed tube above.  What is the frequency of the sound, assuming the speed of sound is 343 m/s?

Solution: Seven A-N pairs: 7(l / 4) = 1.5 m   l = 4 (1.5) / 7      = 0.86 m  v = l f  f = v / l    = 343 / 0.86    = 399 Hz

    Fundamental Frequency

Fundamental frequency =   f1 = v / 4L

l = v / f      = (343 m/s) / 400 s-1    = 0.86 m Distance between antinode and node is one-quarter wavelength. l / 4 = L    l = 4L f = v / l = v / 4L   = 343 / 4   = 85.8 Hz

   Harmonics in a Closed Tube
      (L = 34.3 cm)

Node-Antinode  Pattern		l	f = v / l
A--N	l / 4 = L	4 L / 1	f1 = 1 (v / 4L)       = 25 Hz
A--N--A--N	3 l  / 4 = L	4 L / 3	f2 = 3 (v / 4L)     = 75 Hz
A--N--A--N--A--N	5 l / 4 = L	4 L / 5	f3 = 5 (v / 4L)          = 125 Hz

   The "harmonics" of an instrument are the allowed standing wave
  oscillations.  The fundamental frequency is the first harmonic.  The
  second harmonic is the next-highest frequency; it's also called the
  "first overtone".

 Optimum Response in Human Ear

Between outer ear and eardrum: L = one inch = 2.54 cm. (A tube closed at one end.) Fundmental frequency occurs when node-antinode pattern is A-N.   Thus, l / 4 = L:   l = 4 L       = 4 (0.0254)       = 0.1016 m    f = (343 m/s) / 0.1016 m)      = 3376 Hz

Human ear responds to 3000 Hz frequency more efficiently than to any other frequency.

  Tubes Open at Both Ends

Antinodes always at open ends of tubes. ------------------------- Nodes and antinodes always alternate. ------------------------ In this exa-mple, an antinode is at the center.  Will this always be true for all frequencies?

 

Does the scope always measure a node?

   Fundamental Frequency of Open-Ended Tubes

Fundamental frequency = f1 = v / 2L

Separation between node and antinode is l / 4 2 (l / 4) = L l = 2 L ------------------------- f  = v / l = v / 2L    = 343 / 2    = 171.5 Hz ------------------------- Note:  the longer the tube, the lower the frequency.

   Demonstrate resonance in open tube of length L = 1.3 m.

   Flutes and Organ Pipes       (Open at both ends)

Low pitch (frequency) =====> High pitch

The longer the tube, the lower the fundamental frequency:   f  = v / 2L Cover all openings; opening one of them gives a different L. The air in the flute initially oscillates with a wide range of frequencies, but only those frequencies which allow standing-waves to be set up will be sustained and amplified. Blowing gently on pipe excites the fundamental mode of vibration more than the other modes.

   Open-Tube Harmonics

Configuration		l	f = v / l
A--N--A     (two pairs)	2 (l / 4) = L	l1= 2 L	f1 = (v / 2L)
A--N--A--N--A (four pairs)	4 (l / 4) = L	l2  = l1 / 2	f2 = 2 f1
A--N--A--N--A--N--A (six pairs)	6 (l / 4) = L	l2 = l1 / 3	f1 = 3 f1

     Compare
    Frequencies

(L = 0.343 m) Closed  (Hz) Open  (Hz)      25     50      75   100    125   150    175   200

Demonstration:  Strike with palm and seal the end of an open tube to create an open-tube resonance.   Repeat, this time remove hand immediately to obtain closed-tube resonance at higher frequency.

 

Frequencies Depend on Speed of Sound The fundamental in a tube open at both ends is  f1 = v / 2L.  In air, the speed of sound is 331 m/s at 0 C; in helium, the speed of sound is 965 m/s at 0 C.  If one inhales a little helium and speaks while exhaling it, the resulting voice will have a higher frequency.   What would happen to the sound in organ pipes if they were blown with carbon dioxide (v = 259 m/s) instead of air?

   If carbon dioxide is used, the pipes would resonate at lower frequencies.

 

Standing Waves on Strings   v = [ F / (m / L) ]1/2     F = tension   f = v / l         -------------------------------------------------- What would happen to the frequency if the tension F were quadrupled?

   Standing Waves on a String

Nodes at both ends of guitar string.   Just as in longitudinal standing waves, consecutive node-antinode pairs are separated by one-quarter wavelength. Fundamental:  NAN 2 (l /4) = L             l = 4 L / 2 = 2 L

Problem:  The tension F in a guitar string is 50 N; the length L is 0.60 meter.  The mass of the string is 10 grams. What are the first four harmonics? --------------------------------------------------------- v = [ F / (m / L) ]1/2 = [ 50 / (0.01 / 0.60) ]1/2                                 = 54.8 m/s Fundamental frequency:  NAN Two pairs NA and AN, each  l / 4 long, together add up to the distance L.

2 (l / 4) = L l = 2 L = 1.2 m f1 = v / l  = 54.8 / 1.2                = 45.7 Hz f2 = 2 f1  = 91.4 Hz     (1st overtone) f3 = 3 f1  = 137.1 Hz   (2d overtone) f4 = 4 f1  = 182.8 Hz   (3d overtone)

  How to Create Particular Harmonics

   Explanation of Resonant Modes on a String

Length of string = L        Speed = v Pulse is inverted on reflection at the wall,   and once again when it returns to the starting point, where the cycle begins again.

Return time:    2 L / v Return frequency: 1 / return time f = v / 2L -------------------------------------------- Fundamental frequency = v/2L -------------------------------------------- Harmonics:       f1  = v / 2L       f2  = 2 f1       f3  = 3 f1       .       .       .      fN  = N f1

   Diffraction    The bending of a wave around an obstacle or edges of an opening.

Water-wave diffraction

If no diffraction occurred, sound would be heard only directly in front of opening.

  The Sine Function

q      (degrees)	sin q
0	0.00
30	0.50
60	0.87
90	1.00

 As q increses,
 sin q increases.

  Diffraction Horn Loudspeaker    (Rectangular slit)   

sin q = l / D If l is large (low frequency), bending angle is large.   If D is small, bending angle is large.

 

Woofers and tweeters sin q = 1.22 l / D D = diameter

Speaker Design Problem sin q = 1.22 l / D A 1000 Hz sound comes from a speaker whose diameter D = 0.200 m.  What should be the diameter of a second speaker for 8000 Hz sound if both frequencies are to have the same dispersion?

Another Speaker Design Problem What should be the diameter of a speaker to allow a dispersion of 70 degrees for sound at 3000 Hz? l = v / f = (343 m/s) / 3000 s-1 = 0.114 m sin q = 1.22 l / D     D = 1.22 l / sin q         = 1.22 (0.114 m) / sin 70         =  0.148 m         = 14.8 cm         = 5.83 inches            (2.54 cm = one inch)

  .Beats       (interference in time)

Every second, the maxima will overlap; these are "beats".  The beat frequency is f2 - f1 = 1001 - 1000 = 1 Hz.

Unequal spacings of combs produces a pattern similar to beats.

 Beats

Beat frequency = f2 - f1  = 442 - 440 = 2 Hz

 Demo:  Use 1000 Hz and 1024 Hz tuning forks to hear 24 Hz beat frequency. 

  

Listener hears average of two frequencies, 441 Hz; this average is heard to go through a   maximum two times each second.

Beat Frequency Math     y1 = A0 cos 2pf1t           y2 = A0 cos 2pf2t          y = y1 + y2 = A0 (cos 2pf1t + cos 2pf2t)                                cos A + cos B = 2 cos[(A-B)/2] cos[(A+B)/2]                                 y = 2A0 cos 2p[(f1-f2)/2]t  cos 2p[(f1+f2)/2] t   ---------------------------------------------------------------------------------------------------   Example:  f1 = 440 Hz    f2 = 442 Hz        y = 2A0 cos 2pt cos 2p(441)t   The term cos 2pt controls the beats and corresponds to the envelope   shown as the broken curve in the diagram below:

   Example:  270 Hz and 260 Hz. The average, 265 Hz, will be heard;
   the beat frequency is 10 Hz, but may be too high to be heard.