Question

Air column in two identical tubes is vibrating. Tube A has one end closed and tube B has both ends open. Neglecting end correction, the ratio of the fundamental frequency of air column in tube A to that in tube B is

• A. 2:1
• B. 4:1
• C. 1:4
• D. 1:2

Answer 1

Answer: (D)

1:2

Answer 2

For a closed-end tube:
fA = $\frac {v}{4L}$
For an open-end tube:
fB = $\frac {v}{2L}$
Where fA and fB are the fundamental frequencies of Tube A and Tube B, v is the speed of sound in air, and L is the length of the tubes.
We are given that the tubes are identical, which means their lengths are the same (LA = LB = L).
Taking the ratio of fA to fB, we have:
$\frac {f_A}{f_B}$ = $\frac {v/4L}{v/2L}$
$\frac {f_A}{f_B}$ = $\frac {v}{4L}$ x $\frac {2L}{v}$
$\frac {f_A}{f_B}$ = $\frac {1}{2}$
Therefore, the ratio of the fundamental frequency of the air column in Tube A to that in Tube B is 1:2, or option (D) 1:2.