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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    x and y displacements of a particle are given as x(t)=a \[\sin \omega t\] and \[y(t)=a\sin 2\omega t.\] Its trajectory will look like : JEE Main Online Paper (Held On 10 April 2015)

    A)

    B)

    C)

    D)

    Correct Answer: B

    Solution :

                     \[x=a\sin wt\]                  \[y=a\sin 2wt\] \[y=2a\sin wt\cos wt\]  \[y=2x\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}\] \[y=\frac{2}{a}x\sqrt{\left( a-x \right)\left( a+x \right)}\]


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