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Haryana PMT Haryana PMT Solved Paper-2006

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

    A)                  7.2 m                                   

    B)  4 cm

    C)                  6 cm                                     

    D)  0.72m

    Correct Answer: A

    Solution :

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


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