Question

The bob of a swinging seconds pendulum (one whose time period is $2\, s$ ) has a small speed $v_{0}$ at its lowest point. Its height from this lowest point $2.25 \,s$ after passing through it is given by

• A. $\frac{v_{0}{ }^{2}}{2 g}$
• B. $\frac{v_{0}{ }^{2}}{g}$
• C. $\frac{v_{0}{ }^{2}}{4 g}$
• D. $\frac{9 v_{0}{ }^{2}}{4 g}$

Answer 1

Answer: (C)

$T =2 \sec$.
at $t=T+T / 8$
$v=A \omega \cos \omega t=\frac{v_{0}}{\sqrt{2}}$
By Mechanical energy conservation:
$\left(\frac{v_{0}}{\sqrt{2}}\right)^{2}=v_{0}^{2}-2\, g h$
$h=\frac{v_{0}^{2}}{4 g}$