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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    A particle executing a simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

    A)  \[\frac{3}{2}s\]                                                

    B)  \[\frac{1}{2}s\]

    C)  \[\frac{3}{4}s\]                                                

    D)  \[\frac{1}{4}s\]

    Correct Answer: B

    Solution :

    Equation for simple harmonic motion                 \[y=a\sin \left( \frac{2\pi }{T} \right)\,t\]                 \[\frac{a}{2}=a\sin \left( \frac{2\pi }{T} \right)\,t\] \[\Rightarrow \]               \[\frac{2\pi }{T}\,t=\frac{\pi }{6}\]                            \[\left( \because \,\,y=\frac{a}{2} \right)\]                 \[t=\frac{T}{12}=\frac{6}{12}=\frac{1}{2}s\]


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