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Haryana PMT Haryana PMT Solved Paper-2002

  • question_answer
    A particle of mass 50 g executes SHM with a        time period of 0.18 sec. The amplitude of             vibration is 10 cm, the maximum force on the particle is around :

    A)  10 N                                     

    B)  6 N

    C)  30 N                                     

    D)  35 N

    Correct Answer: B

    Solution :

                    Maximum force, \[F=m{{\omega }^{2}}a\] \[=m{{\left( \frac{2\pi }{T} \right)}^{2}}a=\frac{4{{\pi }^{2}}ma}{{{T}^{2}}}\] \[=\frac{4\times {{(3.14)}^{2}}\times 50\times {{10}^{-3}}\times 10\times {{10}^{-2}}}{{{(0.18)}^{2}}}=6N\]


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