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J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    A particle moves so that its acceleration a is given by \[a=-bx,\] where x is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

    A)  \[2\pi \sqrt{b}\]        

    B)  \[\frac{2\pi }{b}\]

    C)  \[\frac{2\pi }{\sqrt{b}}\]

    D)  \[2\frac{\pi }{\sqrt{b}}\]             

    Correct Answer: C

    Solution :

    The relation between acceleration  and displacement (x) for a body in SHM is \[a=-{{\omega }^{2}}x\] Given,        \[a=-bx\] On comparing the two equations, we get \[{{\omega }^{2}}=b\] \[\therefore \] \[\omega =\sqrt{b}\] Since,     \[\omega =\frac{2\pi }{T}\] \[\therefore \] \[\frac{2\pi }{T}=\sqrt{b}\] \[\Rightarrow \] \[T=\frac{2\pi }{\sqrt{b}}\]


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