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

  • question_answer
    A particle in S.H.M. is described by the displacement function \[x(t)=a\cos (\omega t+\theta )\]. If the initial \[(t=0)\] position of the particle is 1 cm and its initial velocity is \[\pi \,cm/s\]. The angular frequency of the particle is \[\pi \,rad/s\], then it?s amplitude is                                                                  [AMU (Med.) 2002]

    A)            1 cm                                         

    B)            \[\sqrt{2}\,cm\]

    C)            2 cm                                         

    D)            2.5 cm

    Correct Answer: B

    Solution :

                       \[x=a\cos (\omega \,t+\theta )\]                                                   ?.(i)    and  \[v=\frac{dx}{dt}=-a\omega \sin (\omega \,t+\theta )\]        ?.(ii)                    Given at \[t=0\], \[x=1\,cm\] and \[v=\pi \] and \[\omega =\pi \]                    Putting these values in equation (i) and (ii) we will get \[\sin \theta =\frac{-1}{a}\] and \[\cos \theta =\frac{1}{A}\] Þ \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ={{\left( -\frac{1}{a} \right)}^{2}}+{{\left( \frac{1}{a} \right)}^{2}}\]Þ \[a=\sqrt{2}\,cm\]


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