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Uttarakhand PMT Uttarakhand PMT Solved Paper-2006

  • question_answer
    A particle is executing the motion\[x=a\,\cos (\omega t-\text{ }\!\!\theta\!\!\text{ }).\]\[\text{( }\!\!\omega\!\!\text{ t- }\!\!\theta\!\!\text{ )}\text{.}\]). The velocity of the particle is:

    A)  \[a\omega \,\sin \text{ }\!\!\theta\!\!\text{ }\]         

    B)  \[a\omega \,\]

    C)  \[a\omega \,\,\sin \text{ }\!\!\theta\!\!\text{ }\]          

    D)  none of these

    Correct Answer: B

    Solution :

     Given equation is \[x=a\text{ }cos(\omega t-\theta )\] So, velocity \[\left| \frac{dx}{dt} \right|=a\omega \sin (\omega t-\theta )\] At\[(\omega t-\theta )=90{}^\circ ,\]the velocity will be maximum, therefore the maximum velocity is \[v=a\omega \]


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