Question

A tuning fork of frequency $256Hz$ resonates with a closed organ pipe of length $25.4cm$. If the length of the pipe be increased by $2mm$, then calculate the number of beats produced per second.

Answer 1

$n_1=256=\frac{v}{4l_1}=\frac{v}{4\times 25.4}$

$\therefore v=256\times 4\times 25.4=256\times 101.6$cm/s

$n_2=256=\frac{v}{4l_2}=\frac{256\times 101.6}{4\times 25.6}=254$Hz

Number of beats per second$=n_1-n_2=256-254=2$