Question

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?

Answer 1

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`

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`