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JEE Main & Advanced Physics Simple Harmonic Motion Question Bank Displacement of S.H.M. and Phase

  • question_answer
    A particle executing S.H.M. of amplitude 4 cm and T = 4 sec. The time taken by it to move from positive extreme position to half the amplitude is                                [BHU 1995]

    A)            1 sec                                        

    B)            1/3 sec

    C)            2/3 sec                                    

    D)            \[\sqrt{3/2}\]sec

    Correct Answer: C

    Solution :

                       Equation of motion \[y=a\cos \omega t\] Þ \[\frac{a}{2}=a\cos \omega t\Rightarrow \cos \omega t=\frac{1}{2}\Rightarrow \omega t=\frac{\pi }{3}\]            \[\Rightarrow \frac{2\pi t}{T}=\frac{\pi }{3}\Rightarrow t=\frac{\frac{\pi }{3}\times T}{2\pi }=\frac{4}{3\times 2}=\frac{2}{3}\sec \]


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