Question

The two nearest harmonics of a tube closed at one end and open at the other end are $220\text{Hz}$ and $240\text{Hz}$ respectively. What is the fundamental frequency of the system?

• A. $10\text{Hz}$
• B. $20\text{Hz}$
• C. $30\text{Hz}$
• D. $15\text{Hz}$

Answer 1

The correct option is A $10\text{Hz}$

The tube mentioned in question is a closed organ pipe for which the fundamental frequency is,
$f_1$ $=\frac{v}{4l}$
The frequency of vibration is given by,
$f=(2n-1)\frac{v}{4l}$
Where $n=1,2,\mathrm{3......},\text{etc}$
The difference between two successive or nearest harmonics is,
$\Delta f=\left(2\right[n+1]-1)\frac{v}{4l}-(2n-1)\frac{v}{4l}$
$\Rightarrow \Delta f=\frac{2v}{4l}$
Here, $\Delta f=240-220=20\text{Hz}$
$\Rightarrow \frac{2v}{4l}=20$
$\text{Or},2f_1=20$
$\therefore f_1=10\text{Hz}$
Therefore, fundamental frequency of system is $10\text{Hz}$