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A simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in horizontal position from the wall and is left free to move. The bob hits on the wall the coefficient of restitution is $\frac{2}{{\sqrt 5 }}$. After how many collisions the amplitude of vibration will become less than $60°$
$6$
$3$
$5$
$4$
Solution
(b) From the relation of restitution $\frac{{{h_n}}}{{{h_0}}} = {e^{2n}}$ and
${h_n} = {h_0}(1 – \cos 60^\circ )$
$⇒$ $\frac{{{h_n}}}{{{h_0}}} = 1 – \cos 60^\circ = {\left( {\frac{2}{{\sqrt 5 }}} \right)^{2n}}$
$⇒$ $1 – \frac{1}{2} = {\left( {\frac{4}{5}} \right)^n}$
$⇒$ $\frac{1}{2} = {\left( {\frac{4}{5}} \right)^n}$
Taking log of both sides we get
$\log 1 – \log 2 = n(\log 4 – \log 5)$
$0 – 0.3010 = n(0.6020 – 0.6990)$
$ – \,0.3010 = – n \times 0.097$
==> $n = \frac{{0.3010}}{{0.097}} = 3.1 \approx 3$
