Question

The path length of oscillation of simple pendulum of length 1 metre is 16 cm. Its maximum velocity is $(g = \pi^2 \, m/s^2)$

• A. $2 \pi \, cm/s$
• B. $4 \pi \, cm/s$
• C. $8 \pi \, cm/s$
• D. $16 \pi \, cm/s$

Answer 1

Answer: (C)

$8 \pi \, cm/s$

Answer 2

Given,
Length of the pendulum $(l)=1 m$
$\therefore$ Amplitude $(a)=\frac{\text { Path length }}{2}$
$=\frac{16}{2}=8 cm$
Acceleration due to gravity $(g)=\pi^{2} m / s ^{2}$
We know that, time period $(T)=2 \pi \sqrt{\frac{l}{g}}$
$=2 \pi \sqrt{\frac{1}{\pi^{2}}}$
$T=\frac{2 \pi}{\pi}$
$T=2\,s$
$\therefore$ Maximum velocity $\left(v_{\max }\right)=a \omega$
$= a \times \frac{2 \pi}{T} \left(\because \omega=\frac{2 \pi}{T}\right)$
$= \frac{8 \times 2 \times \pi}{2}=8 \pi cm / s$