Question

A tuning fork vibrates at 264 Hz. If the speed of sound in air is 350 m/s the length of the shortest closed organ pipe that will resonate with the tuning fork is

• A. 13 cm
• B. 23 cm
• C. 33 cm
• D. 43 cm

Answer 1

The correct option is C 33 cm

The resonant frequency of a closed organ pipe of length ‘l’ is $\frac{nv}{4l}$, where ‘n’ is a positive odd integer and ‘v’ is the speed of sound in air. To resonate with the given tuning fork
$\frac{nv}{4l}=264\Rightarrow I=\frac{n\times 350}{4\times 264}$
For ‘l’ to be minimum, n = 1 so that
$l_{min}=\frac{350}{4\times 264}=33cm$