WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    The displacement of a particle in a periodic motion is given by \[y=4{{\cos }^{2}}\left( \frac{t}{2} \right)\sin (1000t).\]This displacement may be considered as the result of superposition of\[n\]independent harmonic oscillations. Here \[n\] is

    A)  1       

    B)  2       

    C)  3       

    D)  4

    Correct Answer: C

    Solution :

     Given, \[Y=4{{\cos }^{2}}\left( \frac{t}{2} \right).\sin (1000t)\] \[=2\times 2{{\cos }^{2}}\left( \frac{t}{2} \right).\sin (1000t)\] \[\left[ \because \,2\,{{\cos }^{2}}\frac{t}{2}=(1+cos\,t) \right]\]\[=2(1+cos\,t)sin\,(1000\,t)\] \[=2\sin (1000t)+2cos\,t.sin(1000t)\] \[=2\sin (1000t)+2sin(1000t).cos\,t\] \[=2\,\,\sin \,(1000t)+sin(1000t\,+t)\] \[+\,\sin \,(1000t-t)\] \[[\because \,2sin\,A.cos\,B=sin(A+B)\,+\sin (A-B)]\] \[=2\,\,\sin (1000t)+\sin (1001t)+sin(999t)\] So, \[n=3\]


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