MCQ
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
  • $\frac{{\pi A\sqrt 3 }}{T}$
  • B
    $\frac{{\pi A}}{T}$
  • C
    $\frac{{\pi A\sqrt 3 }}{{2T}}$
  • D
    $\frac{{3{\pi ^2}A}}{T}$

Answer

Correct option: A.
$\frac{{\pi A\sqrt 3 }}{T}$
a
(a)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}} $

$= \frac{{2\pi }}{T}\sqrt {\frac{{3{A^2}}}{4}}$

$= \frac{{\pi A\sqrt 3 }}{T}$.

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