Question

How is the lowest resonant frequency, $f_c$, for a tube with one closed end related to the lowest resonant frequency, $f_0$, for a tube with no closed ends?

• A. $f_c=\frac{1}{4}{f}_0$
• B. $f_c=\frac{1}{2}{f}_0$
• C. $f_c=f_0$
• D. $f_c=2f_0$
• E. $f_c=4f_0$

Answer 1

The correct option is B $f_c=\frac{1}{2}{f}_0$

Lowest frequency (fundamental frequency) of a closed tube (closed organ pipe) is given by

$f_c=v/4l$ , where l= length of tube ,

and lowest frequency (fundamental frequency) of an open tube (open organ pipe) is given by

$f_o=v/2l$ , where l= length of tube ,

therefore $f_c/f_o=1/2$

or $f_c=f_o/2$