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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

  • question_answer
    A particle is executing a simple harmonic motion of period 2 s. When it is at its extreme displacement from its mean position, it receives an additional energy equal to what it its mean position had. Due to this, in its subsequent motion

    A) Its amplitude becomes \[\sqrt{2}\] times of its initial value  

    B) Its periodic time will be doubled

    C) Its potential energy will decrease                     

    D) There will be no change in its motion

    Correct Answer: A

    Solution :

    [a] \[\frac{1}{2}K{{A}^{2}}=KE\] at mean position. Now, \[K{{E}_{final}}=2KE\] at mean position \[=\frac{1}{2}m{{(\sqrt{2}{{v}_{0}})}^{2}}\] Hence maximum velocity will be \[\sqrt{2}\] times the previous velocity. Hence, new amplitude will also increase \[\sqrt{2}\] times of its initial value.


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