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NEET Sample Paper NEET Sample Test Paper-33

  • question_answer
    A particle executing simple harmonic motion has a time period of 48. After how much interval of time from \[t=0\] will its displacement be half of its amplitude?

    A)  \[\frac{1}{3}S\]              

    B)  \[\frac{1}{4}S\]

    C)  \[\frac{2}{3}S\]            

    D)  \[\frac{1}{6}S\]

    Correct Answer: A

    Solution :

    The displacement equation of SHM is \[y=a\,\sin \omega t'\] Where co is angular velocity \[\left( \omega =\frac{2\pi }{T} \right)\] and a the amplitude Here,         \[y=\frac{9}{2},\,t'=\frac{t}{4}\] \[\frac{a}{2}=a\,\sin \frac{2\pi t}{4}\] or \[\frac{1}{2}=\,\sin \frac{\pi t}{2}\] \[\therefore \] \[\sin \frac{\pi }{6}=\sin \frac{\pi t}{2}\] or          \[\frac{\pi }{6}=\frac{\pi t}{2}\]   or    \[t=\frac{1}{3}S.\]


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