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

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

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

So,

Next harmonics occurs at

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

$\frac{\left\{\left(2n+1\right)-\left(2n-1\right)\right\}v}{4l}=260-220$

$2\left(\frac{v}{4l}\right)=40⇒\frac{v}{4l}=20Hz$

So, fundamental frequency of the system

=$\frac{v}{4l}=20Hz$

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