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Manipal Medical Manipal Medical Solved Paper-2005

  • question_answer
    A particle executes SHM, its time period is 16 s. If it passes through the centre of oscillation then its velocity is 2 m/s at time 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: \[t=2s,\,\upsilon =2m/s\,T=16s\] \[\upsilon =a\omega \cos \omega t\] \[2=a.\frac{2\pi }{16}.\cos \frac{2\pi }{16}.2\] \[\therefore \] \[a=\frac{16\sqrt{2}}{\pi }=7.2m\]


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