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VMMC VMMC Medical Solved Paper-2006

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


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