Question

Two successive resonance frequencies in an open organ pipe are $1954\text{Hz}$ and $2602\text{Hz}$. Find the length of the tube.
[Given, speed of sound in air is $324{\text{ms}}^{-1}$]

• A. $25\text{cm}$
• B. $50\text{cm}$
• C. $75\text{cm}$
• D. $100\text{cm}$

Answer 1

The correct option is A $25\text{cm}$

Modes of vibration of an air column in an open organ pipe are given by
$f_n=\frac{nv}{2L}$

Difference between any two successive frequencies is given by $|f_{n+1}-f_n|=\frac{v}{2L}$
Therefore, $\frac{v}{2L}=|2602-1954|=648\text{Hz}$
From this, by using the given data we get,
$L=\frac{324}{2\times 648}=0.25\text{m}\text{or}25\text{cm}$

Thus, option (a) is the correct answer.