Question

Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. The length of the tube is: (the speed of sound in air is 324 m/s)

• A. 25 cm
• B. 50 cm
• C. 12.5 cm
• D. 10.5 cm

Answer 1

The correct option is A 25 cm

For an organ pipe open at both ends:
$f_1=\frac{v}{2L}$

$f_2=\frac{2v}{2L}2f_1$

$f_2-f_1=\frac{v}{2L}=2592-1944=648$

Given: $v=324$

$\therefore L=\frac{324}{2\times 648}=\frac{1}{4}meter=25cm$