MCQ
A simple pendulum with a bob of mass $‘m’$ oscillates from $A$ to $ C$ and back to $A$ such that $PB$ is $H.$ If the acceleration due to gravity is $‘g’$, then the velocity of the bob as it passes through $B$ is
  • A
    $mgH$
  • $\sqrt {2gH} $
  • C
    $\sqrt {gH} $
  • D
    Zero

Answer

Correct option: B.
$\sqrt {2gH} $
b
(b) At $B$, the velocity is maximum using conservation of mechanical energy

$\Delta PE = \Delta KE$

==> $mgH = \frac{1}{2}m{v^2}$

==> $v = \sqrt {2gH} $

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