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VMMC VMMC Medical Solved Paper-2011

  • question_answer
    A particle executing simple harmonic motion has a time of 4s. After how much interval of time from t = 0 will its displacement be half of its amplitude?

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

    B) \[\frac{1}{2}s\]                 

    C) \[\frac{2}{3}s\]                 

    D)        \[\frac{1}{6}s\]

    Correct Answer: A

    Solution :

                            Time period T = 4 s From equation of SHM \[y=a\sin \frac{2\pi }{T}t\]                 Here,     \[y=\frac{a}{2}\]                 \[\therefore \]  \[\frac{a}{2}=a\sin \frac{2\pi t}{4}\]                                 \[\frac{1}{2}=\sin \frac{\pi t}{2}\]                                 \[\sin \frac{\pi }{6}=\sin \frac{\pi t}{2}\] \[\frac{\pi }{6}=\frac{\pi t}{2}\] \[t=\frac{1}{3}s\]


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