A simple pendulum is hanging from a peg inser | vedclass

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Question

(b) From the relation of restitution $\frac{{{h_n}}}{{{h_0}}} = {e^{2n}}$ and
${h_n} = {h_0}(1 - \cos 60^\circ )$

$&rArr;$ $\frac{{{h_n}}}{{{h_0}}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(b) From the relation of restitution $\frac{{{h_n}}}{{{h_0}}} = {e^{2n}}$ and
${h_n} = {h_0}(1 - \cos 60^\circ )$

$&rArr;$ $\frac{{{h_n}}}{{{h_0}}} = 1 - \cos 60^\circ = {\left( {\frac{2}{{\sqrt 5 }}} ight)^{2n}}$
$&rArr;$ $1 - \frac{1}{2} = {\left( {\frac{4}{5}} ight)^n}$

$&rArr;$ $\frac{1}{2} = {\left( {\frac{4}{5}} ight)^n}$
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 	imes 0.097$

==> $n = \frac{{0.3010}}{{0.097}} = 3.1 \approx 3$

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