The fundamental frequency of a closed tube is the same frequency as the 2nd harmonic of a stretched string. The mass per unit length of the string is 6.10x10^-4 kg/m. The tension in the string is 48.0N...?

What is the ratio of the length of the string to the length of the closed tube Ls/Lt?

1 Answer
Aug 20, 2017

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When a tune of fundamental frequency is produced in one end closed tube of length #l#, then only one node at closed end and only one antinode at the open end are formed as shown in above figure. If #lamda_c# be the wave length of the tune produced then #l=lamda_c/4#

#=>lamda_c=4l#

And the frequency of the tune will be #f_c=V/lamda_c=V/(4l).....[1]#,
where V is the velocity of sound #=340m"/"s#

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When second harmonic is produced in streched string of length #L# then there exist three nodes and two antinodes in the streched string as shown in Second figure. The frequency of this second harmonics #f_s# is given by the following relation

#f_s=p/(2L)sqrt(T/m)#

where

#p->"harmonic formed"=2#

#L->"length of the streched string"#

#T->"tension of the stretched string"=48N#

#m->"mass per unit length "6.1xx10^-4kg"/"m#

By the given condition

#f_c=f_s#

#=>V/(4l)=p/(2L)sqrt(T/m)#

#=>L/l=(2p)/Vsqrt(T/m)#

#=>L/l=(2xx2)/340sqrt(48/(6.1xx10^-4))~~3.3#