2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Wave Mechanics

JEE Main & Advanced Physics Wave Mechanics Question Bank Organ Pipe (Vibration of Air Column)

  • question_answer
    A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude \[{{\rho }_{0}}\] inside the tube. If the atmospheric pressure is \[{{\rho }_{A}},\] then the ratio of maximum and minimum pressure at the closed end of the tube will be                                [UPSEAT 2002]

    A)            \[\frac{({{\rho }_{A}}+{{\rho }_{0}})}{({{\rho }_{A}}-{{\rho }_{0}})}\]            

    B)            \[\frac{({{\rho }_{A}}+2{{\rho }_{0}})}{({{\rho }_{A}}-2{{\rho }_{0}})}\]

    C)            \[\frac{{{\rho }_{A}}}{{{\rho }_{A}}}\]                           

    D)            \[\frac{\left( {{\rho }_{A}}+\frac{1}{2}{{\rho }_{0}} \right)}{\left( {{\rho }_{A}}-\frac{1}{2}{{\rho }_{0}} \right)}\]

    Correct Answer: A

    Solution :

               Maximum pressure at closed end will be atmospheric pressure adding with acoustic wave pressure                    So \[{{\rho }_{\max }}={{\rho }_{A}}+{{\rho }_{0}}\] and \[{{\rho }_{\min }}={{\rho }_{A}}-{{\rho }_{0}}\]                    Thus \[\frac{{{\rho }_{\max }}}{{{\rho }_{\min }}}=\frac{{{\rho }_{A}}+{{\rho }_{0}}}{{{\rho }_{A}}-{{\rho }_{0}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner