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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    The displacement of a damped harmonic oscillator is given by \[x(t)={{e}^{-01.1t}}\cos (10\pi t+\phi ).\]Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to: [JEE Main 10-4-2019 Morning]

    A) 13 s                 

    B) 7 s

    C) 27 s                 

    D) 4 s

    Correct Answer: B

    Solution :

    \[A={{A}_{0}}{{e}^{-0.1t}}=\frac{{{A}_{0}}}{2}\] \[\ln 2=0.1t\]           \[t=10\ln 2=6.93\approx 7\sec \]


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