Question

A tuning fork vibrates at $260\text{Hz}$. Find the length of the shortest closed organ pipe that will resonate with the tuning fork. Given, speed of sound in air is $340\text{m/s}$.

• A. $30\text{cm}$
• B. $3.8\text{cm}$
• C. $33\text{cm}$
• D. $44\text{cm}$

Answer 1

The correct option is C $33\text{cm}$

Modes of vibration of an air column in a closed organ pipe are given by

$f=\frac{(2n+1)v}{4L}\text{where}n=0,1,2,3,\mathrm{4.....}$

when $f\to$ frequency
$L\to$ Length of pipe.

From the data given in the question,

$260\text{Hz}=\frac{(2n+1)v}{4L}$

$\Rightarrow L=\frac{(2n+1)\times 340}{4\times 260}=0.327(2n+1)$

For minimum length, $n$ value is minimum

$\therefore$ For $n=0$ we get,
$L_{min}\approx 0.33\text{m}\text{or}33\text{cm}$