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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.

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 θ0 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 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 weill 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 meet 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 ]

 

 

Posted by

infoexpert24

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