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Explain how overtones are produced in a: (a) Closed organ pipe (b) Open organ pipe |
Answer» Solution :Closed organ pipesLookat thepictureof a clarin e t, showninfirgure. Itisa pipewith oneclosedandthe otherendopen. Ifone endof a pipeis closed , thewavereflectedat thisclosedend is `180^@ `out ofphasewiththeincomingwave. thusthereis nodisplacementof theparticlesat the closedend . THerefore, nodesare formedat theclosedend andantinodes areformed at openend .  (a)No motionof particleswhichleadsto nodesat closedend andantinodesat openand( fundamentalmode)(N-node, A - antinode) Let usconsiderthe simplestmodeofvibrationof the aircolumncalledthe fundamentalmode. Anti- nodeis formedat theopenand andnodeatclosedend .FROMTHE figure.LetLbe thelengthof thetubeand thewavelengthof the waveproduced. forthe fundamentalmodeof vibration,we have ` L = ( lamda_1 )/( 4 )orlamda_1= 4L` The frequencyof thenoteemittedis `f_1 = ( v)/( lamda_1) = ( v)/(4 L)` whichis calledthe fundamentalnote : Teh freqencieshigherfundamentalfrequencycan beproducedbyblowingairstronglyat openend .Suchfrequenciesare calledovertones . THefigure(B)showsthe secondmode of vibration havingtwonodes and twoantinodes ,  ` 4L = 3 lamda _2` `L=(3lamda _2)/(4 )or lamda_2= ( 4L )/(3) ` Thefrequencyfor this, ` f_2 =( V) /( lamda _2 )= ( 3v )/( 4 L)= 3 F_1` is calledfirstover, sincehere, thefrequencyis threetimesthefundamentalfrequencyit iscalledthirdharmonic . The figure(c )showsthirdmodeof vibrationhavingthreenodesand three anti- nodes  We have` 4L = 5 lamda_3` ` L= (5lamda_3) /(4 )orlamda_3= ( 4 L )/( 5 )` THefrequency`f_3 = ( V)/( lamda _3 ) = ( 5v )/( 4 L)= 5 F_1` is calledsecondovertoneand sincen=5here, thisis calledfifthharmonic, hencetheclosedorgranpipehasonlyoddharmonics and frequencyof thenthharmonicis` f_n= ( 2 n +1)f_1`: Thereforethefrequenciesof harmonicare in theratio ` F_1 : f_2: F_3: F_4 :...= 1:3:5:7:.............(3)` |
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