Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    Two particles executes SHM of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. The phase difference between them is

    A) 30°                                         

    B)  60°

    C)  90°                                        

    D)  120°

    Correct Answer: D

    Solution :

    Let two simple harmonic motions are \[v{{(4/3)}^{1/2}}\] and \[v{{(4/3)}^{1/2}}\] In the first \[2\to 1\] \[L{{i}^{+}},H{{e}^{+}}\]\[_{\text{90}}\text{T}{{\text{h}}^{\text{228}}}\] In the second case,\[_{83}\text{B}{{\text{i}}^{212}},\] \[\alpha \]\[\text{ }\!\!\beta\!\!\text{ -}\] \[\text{8}\alpha \text{,7 }\!\!\beta\!\!\text{ }\]\[4\alpha \text{,7 }\!\!\beta\!\!\text{ }\]\[4\alpha \text{,4 }\!\!\beta\!\!\text{ }\]\[4\alpha \text{,1 }\!\!\beta\!\!\text{ }\] \[C{{u}^{64}}\]\[{{B}_{H}}=0.3\times {{10}^{-4}}Wb/{{m}^{2}}\]  \[{{B}_{H}}=0.1\times {{10}^{-5}}.\]\[0.25\times {{10}^{-6}}T\] On solving we get \[0.36\times {{10}^{-6}}T\]or\[0.66\times {{10}^{-8}}T\]i.e.,\[1.2\times {{10}^{-6}}T\]or\[4x{{10}^{3}}\]


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