A simple pendulum performs simple harmonic motion about X=0 with … - Vidyadip

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

✓

Answer

Correct option: A.

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

(a) Velocity of a particle executing S.H.M. is given by$v=\omega \sqrt{a^2-x^2}=\frac{2 \pi}{T} \sqrt{A^2-\frac{A^2}{4}}=\frac{2 \pi}{T} \sqrt{\frac{3 A^2}{4}}=\frac{\pi A \sqrt{3}}{T} \text {. }$

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