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

The Correct Option is D

For a closed-end tube: 
f_A = \(\frac {v}{4L}\) 
For an open-end tube: 
f_B = \(\frac {v}{2L}\)
Where f_A and f_B 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 (L_A = L_B = L). 
Taking the ratio of f_A to f_B, 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.