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 .......... $m/s$
$\sqrt 2$
$\sqrt {9.8}$
$4.43$
$1/\sqrt 2 $
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 .......... $m/s$
$v = \sqrt {2gl(1 - \cos \theta )} = \sqrt {2 \times 9.8 \times 2(1 - \cos 60^\circ )} $$ = 4.43\,m/s$
Other Language