Question

Two successive resonance frequencies in an open organ pipe are $1800\text{Hz}$ and $2600\text{Hz}$. Find the length of the tube if the speed of sound in air is $340\text{m/s}$.

• A. $20\text{cm}$
• B. $20.5\text{cm}$
• C. $21.25\text{cm}$
• D. $22.25\text{cm}$

Answer 1

The correct option is C $21.25\text{cm}$

Modes of vibration in an open organ pipe are given by
$f_n=\frac{nv}{2L}$ where $n=1,2,\mathrm{3...}$
So for $n^{th}$ harmonic
$f_n=\frac{nv}{2L}$
and for $(n+1{)}^{th}$ harmonic
$f_{n+1}=\frac{(n+1)v}{2L}$
From question data,
$f_{n+1}-f_n=2600-1800$
$\Rightarrow \frac{(n+1)v}{2L}-\frac{nv}{2L}=800$
$\Rightarrow \frac{v}{2L}=800$
$\Rightarrow L=\frac{v}{1600}=\frac{340}{1600}=21.25\text{cm}$