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A simple pendulum consisting of a light inextensible string of length $\ell$ attached to a heavy small bob of mass $m$ is at rest. The bob is imparted a horizontal impulsive force which gives it a speed of $\sqrt{4 g \ell}$. The speed of the bob at its highest point is ( $g$ is the accelaration due to gravity)
$0$
$\sqrt{\frac{1}{3} g \ell}$
$\sqrt{\frac{2}{3} g \ell}$
$\sqrt{\frac{8}{27} g \ell}$
Solution
(D)
$u =\sqrt{ ng \ell}$
$v =\sqrt{4 g \ell}$
$L =\sqrt{5 g \ell}$
Angle of slack is $\theta$ with vertical
Using energy conservation
$\Rightarrow \frac{1}{2} m u^2=\frac{1}{2} m v^2+m g \ell(1+\cos \theta) \ldots \text { (i) }$
and $m g \cos \theta+7=\frac{m v^2}{\ell} \quad \ldots$ $(ii)$
By $(i)$ and $(ii)$ we get
$\left[\cos \theta=\frac{n-2}{3}\right]=\frac{2}{3}$
$\Rightarrow \frac{2}{3} m g=\frac{ mv ^2}{\ell}$ using $\ldots(ii)$
$v =\sqrt{\frac{2}{3} g \ell}$
At $H _{\max } v ^{\prime}= v \cos \theta=\sqrt{\frac{2}{3} g \ell} \times \frac{2}{3}= v ^{\prime}=\sqrt{\frac{8}{27} g \ell}$
