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Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    An SHM is given by  y = 5 [sin (3\[\pi \]t) + \[\sqrt{3}\]cos (3\[\pi \]t)]                 What is the amplitude of the motion of y in         metre?

    A)                  10                                         

    B)  20

    C)                  1                                            

    D)  5

    Correct Answer: A

    Solution :

                    Given equation is                 \[y\,=\,5\,[\sin \,3\,\pi t\,+\,\sqrt{3\,\cos \,3\,\pi t}]\]                 or            \[y\,=\,5\times 2\,\left[ \sin \,3\,\pi t\times \frac{1}{2}=\frac{\sqrt{3}}{2}\cos \,3\,\pi t \right]\]                 or            \[y\,=\,10\,\sin \,\left( 3\pi t\frac{\pi }{3} \right)\]


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