Question

Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is
A. 1:2
B. 1:4
C. 2:1
D. 4:1

Answer 1

Hint: We assume the speed of the sound in both the cases for tube A and tube B is constant. Then at the closed end there will be nodes and at the open end there will be anti-node. By using the fundamental frequency formula, we find the ratio of the fundamental frequencies of the air column in both cases.

Formula used:
\[{\nu _{0C}} = \dfrac{v}{{2l}}\]
where \[{\nu _{0C}}\] is the fundamental for the air column with both end open and length l.
\[{\nu _0}_O = \dfrac{v}{{4l}}\]
where \[{\nu _{0O}}\] is the fundamental frequency for the air column with one end open and length l.

Complete step by step solution:
Let the length of the tube A is \[{l_A}\] and the speed of sound wave is v, then the fundamental frequency for the tube A which is open at both the ends will be,
\[{\nu _{0A}} = \dfrac{v}{{2{l_A}}}\]
Let the length of the tube B is \[{l_B}\] and the speed of sound wave is v, then the fundamental frequency for the tube B which is closed at one of the ends will be,
\[{\nu _{0B}} = \dfrac{v}{{4{l_A}}}\]
It is given that both the tubes are identical, so the length of both the tubes will be equal.
\[{l_A} = {l_B}\]

We need to find the ratio of the fundamental frequencies of tube A to tube B.
\[\dfrac{{{\nu _{0A}}}}{{{\nu _{0B}}}} = \dfrac{{\left( {\dfrac{v}{{2{l_A}}}} \right)}}{{\left( {\dfrac{v}{{2{l_B}}}} \right)}} \\ \]
\[\Rightarrow \dfrac{{{\nu _{0A}}}}{{{\nu _{0B}}}} = \left( {\dfrac{2}{1}} \right) \times \left( {\dfrac{v}{v}} \right) \times \left( {\dfrac{{{l_B}}}{{{l_A}}}} \right) \\ \]
\[\therefore \dfrac{{{\nu _{0A}}}}{{{\nu _{0B}}}} = \dfrac{2}{1}\]
Hence, the required ratio of the fundamental frequencies of tube A to tube B is 2:1.

Therefore, the correct option is C.

Note: When the tube is closed or open, it is the representation kind of boundaries the sound wave encounters. If the boundary is flexible then it will be the point of the antinode and if it is rigid then the boundary will be the point of the node.