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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    The equation of a wave is given by y = 10 sin \[\left( \frac{2\pi }{45}t+\alpha  \right).\]If the displacement is 5 cm at t = 0, then the total phase at t = 7.5 sis

    A)  \[\frac{\pi }{3}\]

    B)  \[\frac{\pi }{2}\]

    C)  \[\frac{\pi }{6}\]

    D)  \[\pi \]

    Correct Answer: B

    Solution :

    \[y=10\sin \left[ \frac{2\pi }{45}t+\alpha  \right]\] \[y=5cm\,at\,t=0\] \[\therefore \] \[5=10(sin\alpha )\] or \[sin\alpha =\frac{1}{2}\] or \[\alpha =\frac{\pi }{6}\] Again if t =7.5s Then total phase \[=\frac{2\pi }{45}\times \frac{15}{2}+\frac{\pi }{6}\] \[=\frac{\pi }{3}+\frac{\pi }{6}\] \[=\frac{2\pi +\pi }{6}+\frac{3\pi }{6}=\frac{\pi }{2}\]


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