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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    A particle of mass 10 g is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of (\[\text{ }\!\!\pi\!\!\text{ /5}\]) second. The maximum value of the force acting on the particle is

    A)  25 N

    B)                         5 N

    C)  2.5 N                                    

    D)  0.5 N

    Correct Answer: D

    Solution :

    Force in a simple harmonic motion \[F=mf=m(-{{\omega }^{2}}x)=-m{{\omega }^{2}}A\,\sin \,\omega t\] For maximum value of F, \[\sin \,\,\omega t=1\] \[\therefore \]  \[{{F}_{\max }}=m{{\omega }^{2}}A\]                 \[=m{{\left( \frac{2\pi }{T} \right)}^{2}}A=\frac{m\,\,4{{\pi }^{2}}A}{{{T}^{2}}}\]                 \[=\frac{10\times {{10}^{-3}}\times 4{{\pi }^{2}}\times 0.5}{{{\pi }^{2}}/25}=0.5N\]


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