2. Solved Papers
  3. NEET
  4. Physics
  5. Simple Harmonic Motion

NEET Physics Simple Harmonic Motion NEET PYQ-Simple Harmonic Motion

  • 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 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 :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner