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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Oscillations

  • question_answer
    A small ball of density \[4{{\rho }_{0}}\] is released from rest just below the surface of a liquid. The density of liquid increases with depth as \[\rho ={{\rho }_{0}}(1+ay)\] where \[a=2{{m}^{-1}}\]is a constant. Find the time period of its oscillation. (Neglect the viscosity effects).

    A) \[\frac{2\pi }{\sqrt{5}}\sec \]

    B) \[\frac{\pi }{\sqrt{5}}\sec \]

    C) \[\frac{\pi }{2\sqrt{5}}\sec \]

    D) \[\frac{3\pi }{2\sqrt{5}}\sec \]

    Correct Answer: A

    Solution :

    [a] In equilibrium position \[mg={{F}_{B}}=\frac{m}{4{{\rho }_{0}}}{{\rho }_{0}}(1+a{{y}_{0}})g\Rightarrow 4=1+a{{y}_{0}}\] \[{{y}_{0}}=\frac{3}{2}=1.5\] Now displace it downward by \[\Delta y\] \[\therefore \,\,F=mg-\frac{m}{4{{\rho }_{0}}}{{\rho }_{0}}[1+a({{y}_{0}}+\Delta y)]g\] \[=mg-\frac{mg}{4}[1+a{{y}_{0}}+a\Delta y]\] \[F=-\frac{mga}{4}\Delta y;\] Acceleration \[=-\left( \frac{ag}{4} \right)\Delta y\] \[\therefore \,\,T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{4}{ag}}=\frac{2\pi }{\sqrt{5}}\sec .\]


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