Conservation of mechanical energy

Question

A pendulum has a length l. Its bob is pulled aside from its equilibrium position through an angle $\mathrm{\alpha}$ and then released. The speed of the bob when it passes through the equilibrium position is given by

Moderate

Solution

As shown in Fig., the height attained by the bob when the string subtends an angle $\mathrm{\alpha}$ with the vertical is

$\mathrm{h}=\mathrm{l}-\mathrm{lcos}\mathrm{\alpha}=\mathrm{l}(1-\mathrm{cos}\mathrm{\alpha})$

Its potential energy at the highest point A = mgh, Where m is the mass of the bob. Let v be the speed of the bob when it passes through O. Its kinetic energy at $\mathrm{O}=\frac{1}{2}{\mathrm{mv}}^{2}$. From the principle of conservation of energy, we have

$\frac{1}{2}{\mathrm{mv}}^{2}=\mathrm{mgh}$

$\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\mathrm{gl}(1-\mathrm{cos\alpha})}$

Hence the correct choice is (c).

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