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MGIMS WARDHA MGIMS WARDHA Solved Paper-2009

  • question_answer
    The motion of a particle executing SHM is given by \[x\] = 0.01 sin 100\[\pi \] (+0.05), where \[x\] is in metre and r in second. The time period of motion (in second) is

    A)  0.01                                      

    B)  0.02

    C)  0.1                                        

    D)  0.2

    Correct Answer: B

    Solution :

                     \[x=0.01\sin 100\pi (t+0.05)\]                  ...(i) Standard equation is \[x=A\sin (\omega t+\phi )\]            ...(ii) Comparing Eqs. (i) and (ii), we get \[\omega =100\pi \] \[\therefore \]  \[T=\frac{2\pi }{\omega }=\frac{2\pi }{100\pi }=\frac{1}{50}s\] \[T=0.02s\]


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