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JEE Main & Advanced JEE Main Paper (Held On 19 May 2012)

  • question_answer
    The displacement \[y(t)=A\sin (\omega t+\phi )\]of a pendulum for \[\phi =\frac{2\pi }{3}\] is correctly represented by     JEE Main  Online Paper (Held On 19  May  2012)

    A)                      

    B)                       

    C)                                                                   

    D)                       

    Correct Answer: A

    Solution :

                    Displacement \[y(t)=A\sin (wt+\phi )\][Given] For \[\phi =\frac{2\pi }{3}\]at\[t=0;y=A\sin \phi =A\sin \frac{2\pi }{3}\] \[=A\sin {{120}^{o}}=0.87A[\because \sin {{120}^{o}}\simeq 0.866]\] Graph (a) depicts y = 0.87.4 at t = 0


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