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A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time period  T. The speed of the pendulum at x = \frac{A}{}2 will be 

  • Option 1)

    \frac{\pi A\sqrt{3}}{2}

  • Option 2)

    \frac{\pi A}{T}

  • Option 3)

    \frac{\pi A \sqrt{3}}{2T}

  • Option 4)

    \frac{3\pi^{2} A}{T}

 

Answers (1)

best_answer

Velocity of a particle executing S.H.M is given by 

v = \omega\sqrt{a^{2}-x^{2}} = \frac{2\pi}{T} \sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{2\pi}{T} \sqrt{\frac{3A^{2}}{4}} = \frac{\pi A\sqrt{3}}{T}

 

Simple harmonic as projection of circular motion -

- wherein

x= A\cos \omega t

y= A\sin wt

 

 

 


Option 1)

\frac{\pi A\sqrt{3}}{2}

This is correct.

Option 2)

\frac{\pi A}{T}

This is incorrect.

Option 3)

\frac{\pi A \sqrt{3}}{2T}

This is incorrect.

Option 4)

\frac{3\pi^{2} A}{T}

This is incorrect.

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Aadil

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