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AMU Medical AMU Solved Paper-1997

  • question_answer
    A particle in S.H.M. is described by the displacement function \[x(t)=A\cos (\omega t+\phi ),\omega =\frac{2\pi }{T}\] If the initial\[(t=0)\]position of the particle is 1 cm, its initial velocity is n\[cm\text{ }{{s}^{-1}}\]and its angular frequency is\[n\text{ }{{s}^{-1}},\]then the amplitude of its motion is

    A)  \[\pi \,cm\]              

    B)  \[2\,cm\]

    C)  \[\sqrt{2}\,cm\]          

    D)  \[1\,cm\]

    Correct Answer: C

    Solution :

    : \[x(t)=A\cos (\omega t+\phi )\] where\[\omega =\frac{2\pi }{T}\] At\[t=0,x(t)=1\] \[I=A\cos \phi \]                           .... (i) velocity\[v=\frac{dx}{dt}\] or \[v=-A\sin (\omega t+\phi )\times \omega \] or \[v=-\omega A\sin (\omega t+\phi )\] At\[t=0,\]velocity \[=\pi \] \[\therefore \] \[\pi =-\omega A\sin \phi \] \[\pi =-(\pi )A\sin \phi \] or \[A\sin \phi =-1\]                          ....(ii) Square and add, \[{{A}^{2}}{{\cos }^{2}}\phi ={{A}^{2}}{{\sin }^{2}}\phi ={{(1)}^{2}}+{{(-1)}^{2}}\] \[{{A}^{2}}=2\]            or  \[A=\sqrt{2}cm\]


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