JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    The frequency of oscillation is  \[\left( \frac{10}{\pi } \right)\] of a. particle of mass 0.1 kg which executes SHM along \[x-axis.\]The kinetic energy is 0.3J and potential energy is 0.2 J at position \[x.\]The potential energy is zero at mean position. Find the amplitude of oscillations (in metres):

    A)  \[\frac{1}{2\sqrt{10}}\]            

    B)  \[\frac{1}{\sqrt{10}}\]

    C)   \[\sqrt{10}\]                            

    D)  \[2\sqrt{10}\]

    Correct Answer: A

    Solution :

    \[\omega \,=2\pi f=\,{{\left( \frac{K}{m} \right)}^{1/2}}\] \[K={{(2\pi f)}^{2}}m\] \[\frac{1}{2}\,K{{A}^{2}}=\] total energy of oscillation = 0.5 J \[A=\sqrt{\frac{1.0}{K}}\,=\frac{1}{2\pi f}\,\sqrt{\frac{1.0}{0.1}}\,=\frac{1}{2\sqrt{10}}\]


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