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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The instantaneous displacement of a simple harmonic oscillator is given by \[y=A\cos \left( \omega t+\frac{\pi }{4} \right)\]Its speed will be maximum at the time:

    A)  \[\frac{2\pi }{\omega }\]                            

    B)         \[\frac{\omega }{2\pi }\]

    C)  \[\frac{\omega }{\pi }\]                               

    D)         \[\frac{\pi }{4\omega }\]

    E)  \[\frac{\pi }{\omega }\]

    Correct Answer: D

    Solution :

    Speed is maximum when \[y=A\] \[\therefore \]  \[A=A\cos \left( \omega t+\frac{\pi }{4} \right)\] \[\Rightarrow \]               \[\cos \left( \omega t+\frac{\pi }{4} \right)=1\] \[\Rightarrow \]               \[\omega t+\frac{\pi }{4}=0\] \[\Rightarrow \]               \[t=-\frac{\pi }{4\omega }\]


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