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CMC Medical CMC-Medical Ludhiana Solved Paper-2012

  • question_answer
    If for a particle of mass 10 g executing SHM along a straight line, the time period is 2 s and amplitude is 10 cm then what will be its kinetic energy when it is at 5 cm from its equilibrium position?

    A)  \[0.375\,{{\pi }^{2}}\,\text{erg}\]           

    B)  \[3.75\,{{\pi }^{2}}\,\text{erg}\]

    C)  \[37.5\,{{\pi }^{2}}\,\text{erg}\]              

    D)  \[375\,{{\pi }^{2}}\,\text{erg}\]

    Correct Answer: D

    Solution :

                    Kinetic energy of a particle executing SHM is \[K=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})\] \[=\frac{1}{2}\times 10{{\left( \frac{2\pi }{2} \right)}^{2}}[{{(10)}^{2}}-{{(5)}^{2}}]\] \[=375\,{{\pi }^{2}}\,\text{erg}\]


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