RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    A particle of mass 0.10 kg is executing simple harmonic motion at the rate of 20 oscillation/\[{{s}^{2}}\] and its amplitude is 0.05 m. Its energy at equilibrium position will be

    A)  2 J               

    B)  4 J            

    C)  1 J               

    D)  zero

    Correct Answer: A

    Solution :

     \[E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] \[=\frac{1}{2}\times 0.1\times {{(0.05)}^{2}}\times 4{{\pi }^{2}}\times {{(20)}^{2}}\] \[=1.972J\] \[E=2J\]


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