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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 $large {v^1} = ,{e^n}v$

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

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

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

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