NEET NEET SOLVED PAPER 2019

  • question_answer
    The displacement of a particle executing simple harmonic motion is given by\[y={{A}_{0}}+A\,sin\omega t+B\,cos\omega t\]. Then the amplitude of its oscillation is given by: [NEET 5-5-2019]

    A) \[\sqrt{A_{0}^{2}+{{(A+B)}^{2}}}\]            

    B) A+B

    C) \[{{A}_{0}}=\sqrt{{{A}^{2}}+{{B}^{2}}}\]

    D)   \[\sqrt{{{A}^{2}}+{{B}^{2}}}\]

    Correct Answer: D

    Solution :

    \[y={{A}_{0}}+A\text{ }sin\omega t+B\text{ }cos\omega t\] Amplitude of \[A\text{ }sin\omega t+B\text{ }cos\omega t=\sqrt{{{A}^{2}}+{{B}^{2}}}\]


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