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?

Question

Aicharya · Accepted Answer

Two successive frequencies of closed pipe

\frac{n v}{4 l} = 220

....(i)

\frac{(n + 2) v}{4 l} = 260

......(ii)
Dividing (ii) by (i), we get

\frac{n + 2}{n} = \frac{260}{220} = \frac{13}{11}

11 n + 22 = 13 n

n = 11

So,

11 \frac{v}{4 l} = 220

\frac{v}{4 l} = 20

So fundamental frequency is 20 Hz.

The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is…