A pendulum has a length of 0.4 m and a maxim | vedclass

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(D) Given: Length $L = 0.4 \ m$,maximum speed $v_{max} = 4 \ m/s$ at the lowest point.When the string makes an angle of $30^{\circ}$ with the horizont

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$2 \sqrt{3} \ m/s$

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(D) Given: Length $L = 0.4 \ m$,maximum speed $v_{max} = 4 \ m/s$ at the lowest point.When the string makes an angle of $30^{\circ}$ with the horizontal,it makes an angle of $	heta = 60^{\circ}$ with the vertical.The height $h$ of the pendulum bob above the lowest point is given by $h = L - L \cos 	heta = L(1 - \cos 60^{\circ})$.$h = 0.4(1 - 0.5) = 0.4 	imes 0.5 = 0.2 \ m$.Using the law of conservation of energy between the lowest point and the point at angle $	heta$:$\frac{1}{2} m v_{max}^{2} = \frac{1}{2} m v^{2} + mgh$$v^{2} = v_{max}^{2} - 2gh$$v^{2} = (4)^{2} - 2 	imes 10 	imes 0.2$$v^{2} = 16 - 4 = 12$$v = \sqrt{12} = 2 \sqrt{3} \ m/s$.

$A$ pendulum has a length of $0.4 \ m$ and a maximum speed of $4 \ m/s$. When the string makes an angle of $30^{\circ}$ with the horizontal,its speed will be: $\left[\sin \frac{\pi}{6} = \cos \frac{\pi}{3} = 0.5 \text{ and } g = 10 \ m/s^{2}\right]$

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