JEE Main & Advanced Sample Paper JEE Main - Mock Test - 4

  • question_answer
    Kinetic energy of a particle executing simple harmonic motion in straight line is  \[p{{v}^{2}}\] and potential energy is \[q{{x}^{2}},\] where v is speed at distance x from the mean position. It's time period is given by the expression

    A) \[2\pi \sqrt{\frac{q}{p}}\]           

    B) \[2\pi \sqrt{\frac{p}{q}}\]

    C) \[2\pi \sqrt{\frac{q}{p+q}}\]      

    D) \[2\pi \sqrt{\frac{p}{p+q}}\]

    Correct Answer: B

    Solution :

    \[K.E.=\frac{1}{2}m{{\text{v}}^{2}}=p{{\text{v}}^{2}}\Rightarrow m=2p\]
    \[P.E.=\frac{1}{2}k{{x}^{2}}=q{{x}^{2}}\Rightarrow k=2q\]
    Time period \[T=2\pi \,\sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{2p}{2q}}=2\pi \sqrt{\frac{p}{q}}\]


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