A simple pendulum of length 2 m is given a horizontal push throug… - Vidyadip

MCQ

A simple pendulum of length $2 m$ is given a horizontal push through angular displacement of $60^{\circ}$. If the mass of bob is $200$ gram, the angular velocity of the bob will be $($Take acceleration due to gravity $=10\ m / s ^2)$
$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$

A
$2 \sqrt{2} \ rad / s$
B
$3 \sqrt{2} \ rad / s$
✓
$2 \sqrt{2.5} \ rad / s$
D
$3 \sqrt{2.5} \ rad / s$

✓

Answer

Correct option: C.

$2 \sqrt{2.5} \ rad / s$

Given$, \theta=60^{\circ},$
$L=2 m , A B=L \cos \theta$
$h=L-L \cos \theta=L(1-\cos \theta)$
$\text { P.E. }=\text { K.E. }$
$m g L(1-\cos \theta)=\frac{1}{2} m v^2$
$2 \times 10 \times 2\left(1-\frac{1}{2}\right)=\frac{1}{2} v^2$
$v^2=4 \times 10 $
$\Rightarrow v=2 \sqrt{10} m / s$
$\omega=\frac{v}{L}=\frac{2 \sqrt{10}}{2}$
$=2 \sqrt{2.5}\ rad / s $

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