Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    The equation of motion for a body executing SHM is given by\[y=1.5\text{ }sin(10~+\text{ }5)\]. Then the frequency is given by

    A)  5\[\pi \] Hz           

    B)  1.5 Hz

    C)  5 Hz            

    D)  \[\pi \]Hz

    Correct Answer: C

    Solution :

     \[y=1.5\sin (10\pi t+5)\] Comparing with standard equation \[y=a\sin (\omega t+\phi )\] Frequency,   \[n=\frac{\omega }{2\pi }=\frac{10\pi }{2\pi }\]


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