A simple pendulum with a bob of mass m and le | vedclass

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(B) By the principle of conservation of mechanical energy,the potential energy at the initial position equals the kinetic energy at the lowest positio

Vedclass · Accepted Answer

$\sqrt{\frac{1-\cos 	heta_1}{1-\cos 	heta_2}}$

Vedclass · Answer

(B) By the principle of conservation of mechanical energy,the potential energy at the initial position equals the kinetic energy at the lowest position.$m g x(1-\cos 	heta) = \frac{1}{2} m v^2$$v^2 = 2 g x(1-\cos 	heta)$$v = \sqrt{2 g x(1-\cos 	heta)}$For the first position at angle $	heta_1$,$v_1 = \sqrt{2 g x(1-\cos 	heta_1)}$.For the second position at angle $	heta_2$,$v_2 = \sqrt{2 g x(1-\cos 	heta_2)}$.Taking the ratio of the two speeds:$\frac{v_1}{v_2} = \frac{\sqrt{2 g x(1-\cos 	heta_1)}}{\sqrt{2 g x(1-\cos 	heta_2)}}$$\frac{v_1}{v_2} = \sqrt{\frac{1-\cos 	heta_1}{1-\cos 	heta_2}}$

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