Question

The path length of oscillation of simple pendulum of length $1m$ is $16cm$ . Its maximum velocity is $\left(g = \left(\pi \right)^{2} \left(m s\right)^{-2}\right)$

• A. $2\pi \, cms^{- 1}$
• B. $4\pi \, cms^{- 1}$
• C. $8\pi \, cm \, s^{- 1}$
• D. $16\pi \, cms^{- 1}$

Answer 1

Answer: (C)

Path length $= 1 6 \text{ cm}$
$\therefore $ Amplitude $A = 8 \text{ cm}$

Period $T = 2 \pi \sqrt{\frac{l}{g}}$
$= 2 \pi \sqrt{\frac{l}{\pi ^{2}}} = 2 s$
$= 2 \pi \times \frac{1}{\pi } = 2 s$
Maximum velocity $V_{\text{max} =} = a \omega $
$= A \times \frac{2 \pi }{Τ} $
$= 8 \times \frac{2 \pi }{2} $
$= 8 \pi \, c m s^{- 1}$