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Q.

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°

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a

6

b

3

c

4

d

5

answer is B.

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Detailed Solution

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From the relation of restitution $\frac{h_{n}}{h_{0}} = e^{2 n}$ and
$h_{n} = h_{0} (1 - \cos 60 °) \Rightarrow \frac{h_{n}}{h_{0}} = 1 - \cos 60 ° = {(\frac{2}{\sqrt{5}})}^{2 n}$
$\Rightarrow 1 - \frac{1}{2} = {(\frac{4}{5})}^{n} \Rightarrow \frac{1}{2} = {(\frac{4}{5})}^{n}$
Taking log of both sides we get
$\log 1 - \log 2 = n (\log 4 - \log 5)$
$0 - 0 .3010 = n (0 .6020 - 0 .6990)$
$- 0 .3010 = - n \times 0 .097 \Rightarrow n = \frac{0 .3010}{0 .097} = 3 .1 \approx 3$

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