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KVPY Sample Paper KVPY Stream-SX Model Paper-19

  • question_answer
    The energy of a particle executing simple harmonic motion is given by \[E=A{{x}^{2}}+\text{ }B{{v}^{2}}\] where \[x\] is the displacement from mean position \[~x=0\] and v is the velocity of the particle at x then choose the correct statement(s)

    A) Amplitude of SHM. is \[\sqrt{\frac{2E}{A}}\]

    B) Maximum velocity of the particle during S.H.M. is \[\sqrt{EB}\]

    C) Time period of motion is \[2\pi \sqrt{\frac{B}{A}}\]

    D) Displacement of the particle is proportional to the velocity of the particle.

    Correct Answer: C

    Solution :

    Amplitude is obtained for \[v=0\]
    \[\therefore \]Amplitude \[=\sqrt{\frac{E}{A}}\]
    Maximum velocity is obtained for \[x=0\]
    \[{{v}_{\max }}=\sqrt{\frac{\operatorname{E}}{B}}\]
    \[{{v}_{\max }}=\operatorname{amplitude}\times \,\omega \]\[\Rightarrow \]\[\omega =\sqrt{\frac{A}{B}}\Rightarrow T=2\pi \sqrt{\frac{B}{A}}\]


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