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Vibrating Strings and Air Columns

Table of Contents
Introduction
Problems
Answers

      
Introduction

In addition to the relationships shown to the left for strings, remember the following equations we have used earlier.

v = f     also     and 

   
If the strings are uniform, a special equation exist.

L1          f2 
-----  =  -----                 
L2          f1  

   

String instruments use a mechanical amplifier called a sounding board to increase the intensity.  It also effects the quality of the sound.  The best violins are made of wood that has seasoned for several centuries.

The above diagrams show the relationship between L and .  Observe that an open tube has all possible harmonics while the tube with a closed end can only have the odd harmonics.

Remember the velocity of sound is temperature dependent.   v = (331 + 0.60T)  

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Problems

1.      An organ pipe is 155 cm long.  What are the fundamental and first three audible overtones if the pipe is (a) closed on one end, (b) open on both ends.
   
2. A pipe in air at 20.oC is designed to produce two successive harmonics at 240 Hz and 280 Hz.  How long must the pipe be, and is it closed or open?
   
3. The G string of a violin has a fundamental frequency of 196 Hz.  The length of the vibrating portion is 32 cm and has a mass of 0.55 g.  Under what tension must the string be placed?
   
4. A uniform narrow tube 2.10 m long is open at both ends.  It resonates at two successive harmonics of frequencies 275 Hz and 330 Hz.  What is the temperature of the air in the tube?

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Answers

1.      An organ pipe is 155 cm long at 20oC.  What is the fundamental frequency and first three audible overtones if the pipe is (a) closed on one end, (b) open on both ends.
   
   
  v = (331 + 0.60T) = 331 + 0.60 x 20oC = 343 m/s
  

  Closed Pipe  fundamental  

         v          343 m s -1 
  f = -----  =  ----------------- = 55.3 Hz
        4L         4 x 1.55 m
   

  Closed tubes only have odd harmonics

  1st overtone = 3 x 55.3 Hz  =  166 Hz

  2nd overtone = 5 x 55.3 Hz  =  277 Hz

  3rd overtone = 7 x 55.3 Hz  =  387 Hz
   

  Open Pipe fundamental

         v          343 m s -1 
  f = -----  =  ----------------- = 111 Hz
        2L         2 x 1.55 m
   

  Open tubes  have all harmonics

  1st overtone = 2 x 111 Hz  =  222 Hz

  2nd overtone = 3 x 111 Hz  =  333 Hz

  3rd overtone = 4 x 111 Hz  =  444 Hz
   

           
Calculate the velocity at 20oC.

  

Equation for frequency of closed pipe fundamental.

   

As shown in the diagram above, tubes closed on one end, only have odd harmonics.

  

    

    

Equation for frequency of open pipe fundamental.

   

   

As shown in the diagram above, tubes open on both ends have all harmonics.
   

   
2. A pipe in air at 20.oC is designed to produce two successive harmonics at 240 Hz and 280 Hz.  How long must the pipe be, and is it closed or open?
   
   
  280 Hz  -  240 Hz  = 40 Hz  = fundamental open

  We were given the 6th and 7th harmonics if open ended.

  The fundamental would have to be 20 Hz if it is closed on one 

  end.  Since you cannot have even harmonics, the difference

  must be for two harmonics.  This would give you the 12 and 14th

  harmonics.  This is not possible.  Thus the tube is open on

  both ends.

         v                     v        343 m s -1 
  f = -----  =>  L =  ----- = ---------------- =  4.3 m long
        2L                   2f       2 x 40. s -1 
     

           
This is the fundamental if the tube 

is open on both ends.

The difference is equal to 2 times the fundamental if closed on one end.

Open tubes have all harmonics.

Tubes closed on one end have only odd harmonics.

  

Equation for fundamental if open tube.
   

   
3. The G string of a violin has a fundamental frequency of 196 Hz.  The length of the vibrating portion is 32 cm and has a mass of 0.55 g.  Under what tension must the string be placed?
   
   
  = 2L for the fundamental of a string.

  = 2 x 0.32 m = 0.64 m

       

= f

   
            m f22         0.00055 kg x (196 s -1)2 x (0.64 m)2 
  FT = ------------ = ----------------------------------------------------
               L                                 0.32 m

  
  FT  =  27 N

           
Fundamental of a string.

   

   

both are equal to velocity.

   

Solve for Tension.
   

   
4. A uniform narrow tube 2.10 m long is open at both ends.  It resonates at two successive harmonics of frequencies 275 Hz and 330 Hz.  What is the temperature of the air in the tube?
   
    
 Fundamental = 330 Hz  -  275 Hz  =  55 Hz

 v = 2f L  = 2 x 55 s -1 x 2.10 m = 231 m s -1   

 v = 331 + 0.60 T

   
        v  -  331         (231  - 331) m s -1 
T = -------------  =  -------------------------  =  -170oC
          0.60               0.60 oC -1 m s -1  
    

           
Difference in frequencies is the fundamental for open end tubes.

v = 2f L for fundamental of open end tubes.

solve velocity equation for temperature.

 

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