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					                                     Air Column Resonance
                                    The Speed of Sound in Air



     1.        Introduction and Objectives:

              Systems generally have or one or more natural vibrating frequencies. When a system
     is driven at a natural frequency, there is a maximum energy transfer and the vibrational
     amplitude increases to a maximum. For example, when you push a person on a swing, if you
     apply pushes at the proper time (frequency), there will be maximum energy transfer and the
     person will swing higher (greater amplitude). When a system is driven at a natural frequency,
     we say it is in Resonance (with the driving force) and refer to the particular frequency at
     which this occurs as a resonance frequency.

             As seen earlier in the year, a pendulum has only one natural frequency. However, a
     stretched string and an air column can have several natural frequencies depending on the
     number of wavelength segments that can be “fitted” into the system. (Recall when Mr.
     Koenig produced several wavelength segments on the stretched string.)

     From the relationship between the frequency , the wavelength , and the wave speed v, or
      v = 

It can be seen that if the frequency and wavelength are known, the wave speed can be determined.
If the wavelength and wave speed are known, the frequency can be determined.

                          After performing this lab and analyzing the data, you should be able to:

                          1        Explain why a closed organ pipe has particular resonant frequencies
                          2        Tell how the speed of sound in air is affected by temperature and what
                                   affect this has on air column resonance
                          3        Describe how a resonance tube can be used to measure the speed of
                                   sound in air.

     2.        Theory

          Air columns in pipes or tubes of fixed length have particular resonant frequencies. For
          example, in a closed organ pipe (closed at one end) of length L, when the air column is
          driven at a particular frequency, it vibrates in resonance. The interference of the waves
          traveling down the tube and the reflected waves traveling up the tube produces a
          (longitudinal) standing waves, which must have a node at the closed end and an antinode
          (maximum amplitude) at the open end. See diagram.
          The resonance frequencies of a pipe or tube (air column) depend on its length “L”.
          Only a certain number of wavelengths or “loops” can be “fitted” into the tube
          length with the node – antinode requirements. Since each loop corresponds to one
          half wavelengths, resonance occurs when the length of the tube is nearly equal to an
          odd number of quarter wavelengths. i.e. L = /4, L = 3/4, L = 5/4, etc           or
            L = n/4 where n = 1,3,5,… and = 4L/n


2009-10, Koenig Physics                             1
Incorporating the frequency , and the velocity , through the general relationship  =
one derives a formula for resonant frequencies of a air column.


         nv
  fn 
         4L
                 n= 1,3,5,…..




2009-10, Koenig Physics                      2
        When an antinode is at the open end of the tube, a loud resonance tone is heard. Hence,
the tube lengths for antinodes to be at the open end of the tube can be determined by lowering the
water level in the tube and “listening” for successive resonances.

No end correction is needed for the antinode occurring slightly above the end of the tube in
this case, since the differences in tube lengths for successive antinodes is equal to /2.
Ask Mr. Koenig to explain

If the frequency of the driving tuning fork is known and the wavelength is determined by
measuring the difference in tube length between successive antinodes,
         L=/2 or =2L
The speed of sound in air s can be determined from:
         s=

The speed or sound in air is temperature dependent and is given to a good approximation
over the normal temperature range by:

                      Vs= 331.5 + 0.6Tcelsius
                      Temp. correction equation

The equation shows that the speed of sound at 0 0C is 331.5m/s and increases by 0.6m/s for
each degree of temperature increase.

For example, if the air temperature is Tc=20 0C (room temp, 68 0F), then:

Vs = 331.5 + (0.6)(20) = 343.5m/s



2a.       Questions


          1. Suppose that the laboratory temperature were 5 0C higher than the temperature at
               which you performed the experiment. Explain what effect(s) this would have on the
               experimental results. (Remember to think about each variable and what might happen to it.)


Extra credit
          2. With the water in the tube at the level for the tube length L1 of the first resonance of
               the first tuning fork,

                    a) Would another tuning fork with a frequency lower than that of the first
                       tuning fork produce resonance?
                    b) Would another tuning fork of some higher frequency produce a resonance?
                    c) If you answered yes to either case, what would be the other frequency?

                                          (Hint: Look at page 2 for information)




2009-10, Koenig Physics                                        3
     3.        Procedure

     *Work at distance from each lab group. Try to use a frequency of 512HZ or higher.
     Higher frequency shorter wavelength.

         Measure the inside of the resonance tube with a ruler, you are looking for the inside
          diameter of the tube. Record in the data table.

         Place several rubber bands around the air column (optional)


         Record the room temperature in Celsius

         Set up the resonance tube apparatus as demonstrated by Mr. Koenig

         With the water level near the top of the tube there should be little water in the reservoir
          can. If this is not the case remove some water from the can to prevent overflow and
          spilling when the can becomes filled on lowering.

         Practice lowering and raising the water level in the tube to get a “feel” of the apparatus.

         With the water level in the tube near the top, turn on the frequency generator and set it to
          a desired frequency. Be sure to keep the amplitude as low as possible so as not to
          interfere with other groups.


         Keep the speaker slightly above the top of the tube so that the sound is directed
          downward. Be careful not to let the speaker touch the top of the tube.

         With the speaker above the tube, lower the reservoir can. The water level in the tube will
          fall slowly, and successive resonances will be heard.

         As the water passes through successive resonance points mark them with the rubber
          bands. Repeat, moving the water both up and down and adjust the location of the rubber
          bands to correspond to the loudest observed resonance.

         Determine the length from the top of the tube for the first resonance condition and record.
          Repeat this procedure for the other observed resonance positions

         Repeat this process for two other frequencies.

         Compute the average wavelength for each frequency from your data

         In this lab it is not necessary to add to your average wavelength 4 times the tube
          diameter. Ask Mr. Koenig to explain.




2009-10, Koenig Physics                             4
Temperature______________

Inside diameter of tube I.D. _____________Units


                 Frequency, 1_________HZ                              Frequency, 2_________HZ


                      Position                                         Position
                 of resonance (    )        *
                                             L ( )                 of resonance ( )           *
                                                                                                L ( )
         L1
         L2
         L3
                      Average L
                          (L = 1/2)

                          Average
                            (2xL)

Show all work

Vs from frequency 1 data______________________________

Vs from frequency 2 data _______________________

Average Vs ____________________

Vs from using temperature correction equation. _________________________
Percent error______________________




* Due to the length constrictions of the air column apparatus you may be unable to find a L 3 value for frequencies
below 512HZ. Try and use frequencies above 512HZ.




2009-10, Koenig Physics                                    5

				
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