Question

A tuning fork is vibrating at $250\text{Hz}.$ The length of the shortest closed organ pipe that will resonate with the tuning fork will be _________$\text{cm}$.

(Take speed of sound in air as $340{\text{ms}}^{-1}$)

• A. 34.0
• B. 34.00
• C. 34

Answer 1

Answer: (A), (B), (C)

Length of shortest closed organ pipe for resonance is $l=\frac{\lambda }{4}\Rightarrow \lambda =4l$

using,
$v=\lambda f$
$\lambda =\frac{v}{f}$
$4l=\frac{v}{f}$
$l=\frac{v}{4f}=\frac{340}{4\left(250\right)}=0.34\text{m}=34\text{cm}$