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

  • question_answer
    The amplitude of velocity of a particle is given by, \[{{V}_{m}}={{V}_{0}}/(a{{\omega }^{2}}-b\omega +c)\] where \[{{V}_{0}}\], a, b and c are positive: The condition for a single resonant frequency is

    A) \[{{b}^{2}}<4ac\]

    B) \[{{b}^{2}}=4ac\]

    C) \[{{b}^{2}}=5ac\]

    D) \[{{b}^{2}}=7ac\]

    Correct Answer: B

    Solution :

    [b] \[{{V}_{m}}={{V}_{0}}/(a{{\omega }^{2}}-b\omega +c)\]  If there is a single resonant frequency, then this equation should be satisfied for only one that particular resonant frequency, hence \[a{{\omega }^{2}}-b\omega +c=0\] will have equal roots therefore; \[D=0\Rightarrow {{(-b)}^{2}}-4ac=0\Rightarrow {{b}^{2}}=4ac\] particular resonant frequency, hence \[a{{\omega }^{2}}-b\omega +c=0\] will have equal roots therefore;             \[D=0\Rightarrow {{(-b)}^{2}}-4ac=0\Rightarrow {{b}^{2}}=4ac\]


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