Physics, asked by praveen09, 1 year ago

Take a number of test tubes or glass vessels and fill them with water to different levels.Hence the length of air column in the tubes is different .Blow gently across the top of each tube and listen to the pitch of sound produced.
Make a list at what position of the water column the pitch of the sound is maximum.
(need it urgent plz  somebody help)


Anupkashyap: this is an experiment to calculate the resonating frequency..
praveen09: yeah i know

Answers

Answered by kvnmurty
1
     When we blow air across the mouth of the test tube, the air pressure changes across the mouth of the test tube.  The pressure in the air column inside the test tube varies periodically at some specific frequencies.  That frequency depends on the length of the air column.  The pitch and the intensity depend on the frequency.  The pitch is maximum when there is a resonance between our blowing of air and the frequency of pressure changes in air inside test tube.

    If you blow air mildly and steadily, the test tube with longest air column will resonate.  If you blow at high speed,  the test tube with smallest air column will resonate and sound with most intensity.  The air column resonates like a pipe open at one end and closed at the other.

     Suppose the frequency is ∨.  The wavelength of the sound wave is λ.  The speed of sound waves in air is v = 320 meter/sec., at the standard pressure and temperature.   hence,  v =  ∨ * λ.   Also, velocity v is decided by the pressure and density of air as √(γ P/ρ),  γ = ratio of specific heats.

    Let L be the height of air column in the test tube.  The conditions for resonance: are:    L = λ/4,   or    3λ/4    or   5λ/4   or  7λ/4  and so on..

 1.   λ/4 = L,  when there is first harmonic wave inside the test tube.
       So  4 L = λ.   Suppose L = 2 cm.  Then  λ = 8 cm = 0.08 m
          frequency = 320/0.08 = 4 kHz,    Sound of this frequency will resonate.

2.  λ/4 = L  =  4 cm in another test tube.    let us say.  
          λ = 4*4 = 16 cm.    Then  frequency = 320/0.16 = 2 kHz
                    The sound of this frequency will resonate.

Frequency (of fundamental tone) is inversely proportional to the length of air column.           
                    ∨ = v/(4L)


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