The two nearest harmonics of a tube closed at one end and open at other end are $220\text{ Hz}$ and…
The two nearest harmonics of a tube closed at one end and open at other end are $220\text{ Hz}$ and $260\text{ Hz}$. What is the fundamental frequency of the system? [NEET 2017]
(a) $10\text{ Hz}$
(b) $20\text{ Hz}$
(c) $30\text{ Hz}$
(d) $40\text{ Hz}$
Solution
Frequency of $n\text{th}$ harmonic in a closed end tube,
$f = \frac{(2n - 1)v}{4l}$
where, $n = 1, 2, 3, \dots$
Also, only odd harmonics exists in a closed end tube.
Now, given two nearest harmonics are of frequency $220\text{ Hz}$ and $260\text{ Hz}$.
$\therefore \frac{(2n - 1)v}{4l} = 220\text{ Hz}$
Next harmonic occurs at
$\frac{(2n + 1)v}{4l} = 260\text{ Hz}$
On subtracting Eq. (i) from Eq. (ii), we get
$\frac{\{(2n + 1) - (2n - 1)\}v}{4l} = 260 - 220$