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

  • question_answer
    A forced oscillator is acted upon by a force \[F={{F}_{0}}\] sin cot. The amplitude of oscillation is given by              \[\frac{55}{\sqrt{2{{\omega }^{2}}-36\omega +9}}\]. The resonant angular frequency is

    A) 2 unit   

    B) 9 unit

    C) 18 unit

    D) 36 unit

    Correct Answer: B

    Solution :

    [b] At resonance, amplitude of oscillation is maximum \[\Rightarrow 2{{\omega }^{2}}\text{ }-\text{ }36\omega \text{ }+\text{ }9\text{ }is\text{ }minimum\] \[\Rightarrow \,4\omega -36=0\](derivative is zero) \[\Rightarrow \,\omega =9\]


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