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
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
- 2:1
- 4:1
- 1:4
- 1:2
The Correct Option is D
Solution and Explanation
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.
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