A simple pendulum of length \(2 \mathrm{~m}\) 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 \(=10 \mathrm{~m} / \mathrm{s}^2\) ) \(\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)\)

Given: \(l=2 \mathrm{~m}, \theta=60^{\circ}, \mathrm{m}=200 \mathrm{~g}=2 \mathrm{~g}\)


From the figure,
\(\mathrm{T}=\mathrm{mr} \omega^2\)... (i)
also \(\mathrm{T} \cos \theta=\mathrm{mg}\)... (ii)
putting (i) into (ii)
\(\mathrm{mr} \omega^2-\cos \theta=\mathrm{mg}\)... (iii)
putting the given values into equation (iii)
\(\begin{aligned}
& 2 \times 2 \times \omega^2 \frac{1}{2}=2 \times 10 \quad \ldots .\left(\because \cos 60=\frac{1}{2}\right) \\
& \omega^2=10 \\
& \Rightarrow \omega=\sqrt{10} \quad \ldots .(\because \sqrt{10}=\sqrt{2 \times 2 \times 2.5}) \\
& \quad=2 \sqrt{2.5} \mathrm{rad} / \mathrm{s}
\end{aligned}\)