Question 260: Oscillations

Question

The bob of a pendulum of length ' $l$ ' is pulled aside from its equilibrium position through an angle ' $	heta$ ' and then released. The bob will then pass through its equilibrium position with speed ' $v$ ', where ' $v$ ' equal to ( $g=$ acceleration due to gravity)

Accepted Answer

When bob of a pendulum rises up a height ' $h$ ' potential energy at extreme position becomes kinetic energy of mean position.
$$\begin{array}{ll} 
& \mathrm{mgh}=\frac{1}{2} \mathrm{mv}_{\max }^2 \
	herefore \quad & \mathrm{v}_{\max }=\sqrt{2 \mathrm{gh}} \
& l=\mathrm{h}+l \cos 	heta \
	herefore \quad & \mathrm{~h}=l(1-\cos 	heta) \
	herefore \quad & \mathrm{v}_{\max }=\sqrt{2 \mathrm{gl}(1-\cos 	heta)}
\end{array}$$

Answer

$\sqrt{2 g l(1-\cos 	heta)}$

Answer

$\sqrt{2 \mathrm{~g} l(1+\sin 	heta)}$

Answer

$\sqrt{2 g l(1-\sin 	heta)}$

Answer

$\sqrt{2 \mathrm{~g} l(1+\cos 	heta)}$

Oscillations

The bob of a pendulum of length ' l ' is pulled aside from its equilibrium position through an angle ' \theta ' and then released. The bob will then pass through its equilibrium position with speed ' v ', where ' v ' equal to ( g= acceleration due to gravity)

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