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A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Figure). The amplitude is $\theta _{0}$ . The string snaps at $\theta =\frac{\theta _{0}}{2}$ . Find the time taken by the bob to hit the ground. Also, find the distance from A where Bob hits the ground. Assume $\theta_o$ to be small so that $\sin \theta _{0}{\cong } \theta _{0}$ and cos $\theta _{0}{\cong } 1$ .

Answers (1)

Let us consider the diagram at which,

t = 0

$\Theta = \frac{\Theta _{0}}{2} ,\Theta = \Theta _{0} \cos \omega t$

Now, at t = t₁,

$\Theta =\frac{\Theta _{0}}{2}$

Thus, $\frac{\Theta _{0}}{2} = \Theta _{0} \cos \frac{2\pi }{T}t_{1}$

…….. (Given: T = 1sec)

Thus, $\cos 2\pi t_{1} = \frac{1}{2} = \frac{cos \pi}{3}$

$2\pi t_{1} = \frac{\pi}{3}$

Or, $t_{1} = \frac{1}{6}$

$\Theta = \Theta _{0} \cos 2\pi t$

$\frac{d\Theta }{dt} = -\Theta _{0} 2\pi \sin 2\pi t$

Now, at $t = \frac{1}{6}$

$\Theta =\frac{\Theta _{0}}{2}$

$\frac{d\Theta }{dt} = -\Theta _{0} 2\pi \sin 2\pi \frac{1}{6}$

$= -\Theta _{0} 2\pi \frac{3}{\sqrt{2}}$

$= -\Theta _{0} \pi \sqrt{3}$

Now, $\omega= -\Theta _{0} \pi \sqrt{3}$

i.e., $\frac{v}{l}= -\Theta _{0} \pi \sqrt{3}$

Thus, $v= - \sqrt{3}\Theta _{0} \pi l$

By the –ve sign it is clear that the bob’s motion is towards the left.

Now,

$vx = v \cos \frac{\Theta_{0}}{2}$

= $= - \sqrt{3}\Theta _{0} \pi l \cos \frac{\Theta _{0}}{2}$

$vy = v \sin \frac{\Theta _{0}}{2} = - 3\pi \Theta _{0}l \sin \frac{\Theta _{0}}{2}$

Let H’ be the vertical distance covered by vy,

$s = ut + \frac{1}{2} gt^{2}$

$H^{'}=vyt= \frac{1}{2} gt^{2}$

$\frac{1}{2}gt^{2} + 3\pi \Theta _{0}l \sin \frac{\Theta _{0}}{2} \times t - H^{'} = 0$

Thus, $\frac{1}{2}gt^{2} + 3\pi \Theta _{0}l \frac{\Theta _{0}}{2} \times t - H^{'} = 0$ ……. (Given: $\sin \frac{\Theta _{0}}{2} =\frac{\Theta _{0}}{2}$ )

Now, by quadratic formula,

$t = \frac{-B\pm \sqrt{B^{2}-4AC}}{2A}$

$t = \frac{-\frac{\sqrt{3}\pi \theta _{0}^{2}l}{4}\pm \sqrt{(3\pi ^{2}\theta _{0}^{4}l^{2})^{2}-\frac{4.1}{2.g}H^{'}}}{\frac{2g}{2}}$

We will neglect $\theta _{0}^{4} and \theta _{0}^{2}$ since $\theta _{0}$ is very small.

Thus, $t = \frac{+\sqrt{2gH^{'}}}{g } H^{'} = H + H^{"}$

Thus, $\sqrt{\frac{2H}{g}}$

H’’<<H’ as $\frac{\Theta _{0}}{2}$ is very small.

Thus, H = H’

Thus, vxt = distance covered in horizontal

$X =\sqrt{ 3}\pi \theta _{0}l.\sqrt{\frac{2H}{g}}$

…….. (Given: $\cos \frac{\theta _{0}}{2}=1$ )

Thus, $X = \Theta _{0}l\pi \sqrt{\frac{6H}{g}}$

The bob was at a distance of $l\sin \frac{\theta _{0}}{2}=l \frac{\theta _{0}}{2}$ from A at the time of snapping.

Thus, the distance of bob from A where it meets the ground = $l .\frac{\Theta _{0}}{2} - X$

$=l.\frac{\Theta _{0}}{2} -\Theta _{0}l\pi \sqrt{\frac{6H}{g}}$

$= l\Theta _{0}\left [ \frac{1}{2} - \pi\sqrt{ \frac{6H}{g}} \right ]$

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