Haryana PMT Haryana PMT Solved Paper-2003

  • question_answer
    The period of SHM of a particle is 12s. The phase difference between the position at t = 3s and t = 4s will be :

    A)  \[\pi /4\]                            

    B)  \[3\pi /5\]

    C)  \[\pi /6\]                            

    D)  \[\pi /2\]

    Correct Answer: C

    Solution :

                    At, \[t=3s\] \[{{\phi }_{1}}=\omega t=\left( \frac{2\pi }{T} \right)t=\frac{2\pi \times 3}{12}=\frac{\pi }{2}\]                 at            \[t=4s\]                                 \[\phi =\omega t=\left( \frac{2\pi }{T} \right)t\]                                 \[=\frac{2\pi \times 4}{12}=\frac{2\pi }{3}\] Now phase difference                 \[{{\phi }_{2}}-{{\phi }_{1}}=\frac{2\pi }{3}-\frac{\pi }{2}=\pi /6\]


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