First slide

Coefficient of restitution

Question

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°

Difficult

A

6
B

3
C

5
D

4

Solution

From the relation of restitution
$\large \frac{{{h_n}}}{{{h_o}}} = \,{e^{2n}}$
and
$\large {h_n}\, = \,{h_o}\left( {1 - \cos {{60}^0}} \right)\, \Rightarrow \,\frac{{{h_n}}}{{{h_o}}}\, = \,1 - \cos {60^0} = \,{\left( {\frac{2}{{\sqrt 5 }}} \right)^{2n}}\,$
$\large \Rightarrow \,1 - \frac{1}{2}\, = \,{\left( {\frac{4}{5}} \right)^n}\, \Rightarrow \,\frac{1}{2}\, = \,{\left( {\frac{4}{5}} \right)^n}$
Taking log of both sides we ge t
log1–log2 = n(log4–log5)
0–0.3010 = n(0.6020–0.6990)
solving for n $\approx 3$

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