The two nearest harmonics of a tube close at one end and open at other end are 220Hz and 260Hz. What is the fundamental frequency of the system?

(a)10Hz

(b)20Hz

(c)30Hz

(d)40Hz

 

(b)Thinking Process Frequency of nth harmonic in an closed end tube 

                f2n-1v4l               n=1,2,3

Also, only odd harmonics exists in a closed end tube.

Now, given two nearest harmonics are of frequency 220Hz and 260Hz       

So, 2n-1v4l=220Hz      ...(i)

Next harmonics occurs at 

          2n+1v4l=260Hz      ...(ii)

On subtracting Eq. (i) from Eq (ii). we get 

2n+1-2n-1v4l=260-220

2v4l=40v4l=20Hz

So, fundamental frequency of the system 

                    =v4l=20Hz

 

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