NEET Sample Paper NEET Sample Test Paper-9

  • question_answer The displacement of a particle (in metre) from its mean position is given by the equation \[Y=0.2\,\,\,\left( {{\cos }^{2}}\frac{\pi t}{2}-{{\sin }^{2}}\frac{\pi t}{2} \right)\]

    A)  The motion of the above particle is not simple harmonic

    B)  Simple harmonic with the period equal to that of a second's pendulum

    C)  Simple harmonic with the period double that of a second's a pendulum

    D)  Simple harmonic with amplitude \[0.4m\]

    Correct Answer: B

    Solution :

    Given that   \[Y=0.2\,\left( {{\cos }^{2}}\frac{\pi t}{2}-{{\sin }^{2}}\frac{\pi t}{2} \right)\] \[\Rightarrow \]               \[Y=0.2\,\cos \pi t\] This gives,   \[A=0.2,\] \[T=\frac{2\pi }{\pi }=2S\] (as \[\omega =\pi \])

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