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A simple pendulum is hanging from a per inserted in a vertical wall. Its bob is stretched to horizontal position from wall and is left free to move, the bob hits the wall. If coefficient of restitution is $\large \frac{2}{{\sqrt 5 }}$ . After how many collision the amplitude of vibration will become less than 60⁰ (Hint: log2 = 0.3, log5 = 0.7)

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Detailed Solution

If l is length of pendulum, velocity while hitting is given by $\large mgl\, = \,\frac{1}{2}m{v^2}$

$\large v\, = \,\sqrt {2gl}$

After n hittings velocity is $\large {v^1} = \,{e^n}v$

$\large \sqrt {2gl\left( {1 - \cos \theta } \right)} \, \geqslant \,{e^n}\sqrt {2gl}$

$\large 1 \times \frac{1}{2} \geqslant {e^{2n}}\left( {\therefore \theta \, = \,{{60}^0}} \right)$

$\large \frac{1}{2} \geqslant \,{\left( {\frac{4}{5}} \right)^n}\,\,\,\therefore \,e\, = \,\frac{2}{{\sqrt 5 }}\, \Rightarrow \,n\, = 3$

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