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 $60{^\circ}$
A) 6
B) 3
C) 5
D) 4
Explanation
(b) From the relation of restitution $\frac{h_{n}}{h_{0}} = e^{2n}$ and
$h_{n} = h_{0}\left( 1 - \cos{60{^\circ}} \right) \Rightarrow \frac{h_{n}}{h_{0}} = 1 - \cos{60{^\circ}} = \left( \frac{2}{\sqrt{5}} \right)^{2n}$
$\Rightarrow 1 - \frac{1}{2} = \left( \frac{4}{5} \right)^{n} \Rightarrow \frac{1}{2} = \left( \frac{4}{5} \right)^{n}$
Taking log of both sides we get
$\log 1 - \log 2 = n(\log 4 - \log 5)$
$0 - 0.3010 = n(0.6020 - 0.6990)$
$- 0.3010 = - n \times 0.097 \Rightarrow n = \frac{0.3010}{0.097} = 3.1 \approx 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.