KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    A particle executes SHM in a straight line. In the first second starting from rest it travels 'a' distance a and in the next second a distance 'b' in the same direction. The amplitude of S.H.M will be

    A) \[\frac{2{{a}^{2}}}{3a-b}\]

    B) \[a-b\]

    C) \[2a-b\]

    D) \[a/b\]

    Correct Answer: A

    Solution :

    \[x=A\cos \,wt\]  (as it starts from rest at t=0)
    \[A-a=A\cos w\]                         ...(1)
    \[A-(a+b)=A\cos 2w\]                 ...(2)
    Solving (1) and (2) for A we get \[A=\frac{2{{a}^{2}}}{3a-b}\]

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