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The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?
Explanation
Successive odd harmonics differ by twice the fundamental.
Detailed Solution
Two successive frequencies of a closed pipe: $\frac{nv}{4l} = 220$ ...(i) and $\frac{(n + 2)v}{4l} = 260$ ...(ii)
Dividing (ii) by (i): $\frac{n + 2}{n} = \frac{260}{220} = \frac{13}{11}$
$11n + 22 = 13n \Rightarrow n = 11$
So $11\frac{v}{4l} = 220 \Rightarrow \frac{v}{4l} = 20$
So the fundamental frequency is 20 Hz.