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NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • question_answer
    A rectangular block of mass m and area of cross-section A floats in a liquid of density                                                                                    \[\rho\] . If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then:                                                                                                                                       [AIPMT (S) 2006]

    A)                                                                        \[T\propto \,\,\sqrt{\rho }\]

    B)                                                                                    \[T\propto \,\,\frac{1}{\sqrt{A}}\]

    C)  \[T\propto \,\,\frac{1}{\rho }\]

    D)                   \[T\propto \,\,\frac{1}{\sqrt{m}}\]

    Correct Answer: B

    Solution :

    Key Idea: Force applied on the body will be equal to up thrust for vertical oscillations.
    Let block is displaced through \[x\,m,\] then weight of displaced water or upthrust (upwards).
                \[=-Ax\rho g\]
    where A is area of cross-section of the block and \[\rho \] is its density. This must be equal to force \[(=ma)\] applied, where m is mass of the block and a is acceleration.
    \[\therefore \]      \[ma=-Ax\rho g\]
    or         \[a=-\frac{A\rho g}{m}x=-{{\omega }^{2}}x\]
                This is the equation of simple harmonic motion.
                Time period of oscillation
    \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A\rho g}}\]
    \[\Rightarrow \]   \[T\propto \,\frac{1}{\sqrt{A}}\]


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