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AMU Medical AMU Solved Paper-2006

  • question_answer
    A particle executes SHM, its time period is 16 s If its passes through the centre of oscillation then its velocity is 2 m/s at times 2 s. The amplitude will be

    A)  7.2 m                                   

    B)  4 cm

    C)  6 cm                                     

    D)  0.72 m

    Correct Answer: A

    Solution :

                     Given : t = 2s, v = 2 m/s, T = 16 s                 \[v=a\omega \cos \omega t\]                 \[2=a\,.\,\frac{2\pi }{16}.\,\cos \frac{2\pi }{16}.\,\,2\] \[\therefore \]  \[a=\frac{16\sqrt{2}}{\pi }=7.2\,m\]


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