Question

Find out the relation between the lowest resonant frequency, $f_c$, for a tube with one closed end and 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

Answer: (B)

$f_c=\frac{1}{2}{f}_0$

Let the length of the closed tube at one end and open tube at both ends be $L$.

Lowest resonant frequency for tube closed at one end $f_c=\frac{v}{4L}$

Lowest resonant frequency for tube open at both ends $f_o=\frac{v}{2L}$

$\therefore$ $\frac{f_c}{f_o}=\frac{v/4L}{v/2L}=\frac{1}{2}$ $⟹f_c=\frac{f_o}{2}$