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…

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

Aicharya · Accepted Answer

Given: $l=2 \mathrm{~m}, 	heta=60^{\circ}, \mathrm{m}=200 \mathrm{~g}=2 \mathrm{~g}$ From the figure, $\mathrm{T}=\mathrm{mr} \omega^2$... (i) also $\mathrm{T} \cos 	heta=\mathrm{mg}$... (ii) putting (i) into (ii) $\mathrm{mr} \omega^2-\cos 	heta=\mathrm{mg}$... (iii) putting the given values into equation (iii) $\begin{aligned} & 2 	imes 2 	imes \omega^2 \frac{1}{2}=2 	imes 10 \quad \ldots .\left(\because \cos 60=\frac{1}{2}ight) \ & \omega^2=10 \ & \Rightarrow \omega=\sqrt{10} \quad \ldots .(\because \sqrt{10}=\sqrt{2 	imes 2 	imes 2.5}) \ & \quad=2 \sqrt{2.5} \mathrm{rad} / \mathrm{s} \end{aligned}$

A simple pendulum of length $2 \mathrm{~m}$ is given a horizontal push through angular displacement of…

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