MCQ
A pendulum bob on a $2 m$ string is displaced 60 from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path
- A$\sqrt{2} m / s$
- ✓$\sqrt{9.8} m / s$
- C$4.43 m / s$
- D$1 / \sqrt{2} m / s$
✓
Answer
Correct option: B.
$\sqrt{9.8} m / s$
(b) Increment in angular velocity $\omega=2 \pi\left(n_2-n_1\right)$$\omega=2 \pi(1200-600) \frac{ rad }{ min }=\frac{2 \pi \times 600}{60} \frac{ rad }{ s }=20 \pi \frac{ rad }{ s }$
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