
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
163.5k+ views
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.
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.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
