J & K CET Engineering J and K - CET Engineering Solved Paper-2015

  • question_answer
    Two pendula oscillate with a constant phase difference of \[{{45}^{o}}\] and same amplitude. If the maximum velocity of one of them is v and that of other is \['v+x',\] then the value of Y will be

    A)  \[0\]       

    B) \[v/2\]

    C) \[v/\sqrt{2}\]

    D) \[(\sqrt{2})v\]

    Correct Answer: A

    Solution :

    Let velocity of one pendulum is \[{{v}_{1}}={{v}_{0}}\,\,\cos (\omega t+{{\phi }_{1}})\] Velocity of the other pendulum is \[{{v}_{2}}={{v}_{0}}\,\,\cos (\omega t+{{\phi }_{2}})\] According to the question, \[{{\phi }_{2}}-{{\phi }_{1}}={{45}^{o}}\] Clearly, \[|{{v}_{1}}{{|}_{\max }}=|{{v}_{2}}{{|}_{\max }}={{v}_{0}}=v\] Thus,       \[x=0\]


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