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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2010

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

    A)  \[\frac{\pi }{4\omega }\]

    B)  \[\frac{\pi }{2\omega }\]

    C)  \[\frac{\pi }{\omega }\]

    D)  \[\frac{2\pi }{\omega }\]

    Correct Answer: D

    Solution :

     Displacement\[x=A\cos \left( \omega t+\frac{\pi }{4} \right)\] \[v=\frac{dx}{dt}=-A\omega \sin \left( \omega t+\frac{\pi }{4} \right)\] For maximum speed, \[\sin \left( \omega t+\frac{\pi }{4} \right)=1\] Or \[\omega t+\frac{\pi }{4}=\frac{\pi }{2}\] Or \[\omega t=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\] \[t=\frac{\pi }{4\omega }\]


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