Question

A pendulum bob has a speed of $3$ ${ms}^{-1}$ at its lowest position. The pendulum is$0.5$ $m$ long. The speed of the bob, when the length makes an angle of ${60}^{\circ }$ to the vertical will be $(g=10$ $\left(n{s}^{-1}\right)$

• A. $3$ $m{s}^{-1}$
• B. $1/3{ms}^{-1}$
• C. $1/2m{s}^{-1}$
• D. $2m{s}^{-1}$

Answer 1

The correct option is D $2m{s}^{-1}$

Apply energy conservation theorem$,$

energy at lowest position of Bob $=$ energy$,$ when Bob makes $60°$ to the vertical

$1/2m{v}^2=1/2mv{₁}^2+mgl(1-cos60°)$

Here $v$ is speed at Lowest position $,v₁$ is speed $,$ when it makes $60°$ with vertical and $l$ is length of pendulum $.$

$[$Actually, height of Bob $,$ when it makes $60°$ with vertical $=l(1-cos60°)]$

$\therefore v^2=v{₁}^2+2gl(1-cos60°)$

$3^2=v{₁}^2+2\times 10\times 0.5(1-1/2)$

$9=v{₁}^2+5$

$v{₁}^2=4\Rightarrow v₁=2m/s$

So$,$ speed of Bob $=2m/s$

Hence,

option $\left(D\right)$ is correct answer.