Question

A tuning fork vibrates at $264\text{Hz}$. The shortest length of a closed organ pipe that will resonate with the tuning fork is
[Speed of sound in air $=350\text{m/s}$]

• A. $33\text{cm}$
• B. $65\text{cm}$
• C. $75\text{cm}$
• D. $95\text{cm}$

Answer 1

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

We know that for a closed organ pipe, the resonate frequency is given by,
$f=\frac{nv}{4l}$
For the given frequency, and for length to be minimum, $n$ should be minimum i.e.$(n=1)$.
$f=\frac{v}{4l}$
Also, this fundamental frequency will be equal to the frequency of the tuning fork.
$\Rightarrow 264=\frac{350}{4l}$
$\Rightarrow l=\frac{350}{4\times 264}=0.33\text{m}=33\text{cm}$