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AMU Medical AMU Solved Paper-2003

  • question_answer
    The displacement of a particle from its mean position (in metre) is given by \[\text{y }=0.\text{2 sin }\left( \text{10}\pi \text{t }+\text{ 1}.\text{5}\pi  \right)\text{ cos }\left( \text{10}\pi \text{t }+\text{ l}.5\pi  \right)\]The motion of the particle is

    A)  periodic but not SHM

    B)  non periodic

    C)  simple harmonic motion with period 0.1 s

    D)  simple harmonic motion with period 0.2 s

    Correct Answer: C

    Solution :

                     The standard equation of SHM wave of amplitude a, angular frequency co, and phase is                 \[y=a\,\sin (\omega \,t+\phi )\]                ... (i) where t is time. Given equation is \[y=0.2\,\sin \,(10\pi t+5\pi )\,\cos \,(10\pi t+1.5\pi )\] Using the relation \[\sin \,2\theta =2\sin \theta cos\theta \], we have \[y=0.1\,\sin 2\,(10\pi t+1.5\pi )\] \[\Rightarrow \]               \[y=0.1\,\sin \,(20\pi t+3\pi )\]  ... (ii) On comparing Eqs. (i) and (ii), we get                 \[\omega =20\pi \], Hence, time period,                 \[T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }\] \[\Rightarrow \]               \[T=\frac{1}{10}=0.1\,\,s\]          


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