If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0 to `t=tau s`, then `tau` may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with 'b' as the constant of proportionality, the average life time of the pendulum is (assuming dampling is small) in secods :

Options

A. `(2)/(b)`
B. `(0.693)/(b)`
C. b
D. `(1)/(b)`

Answer

A

Explanation

`m(d^(2)x)/(dt^(2))=-kx-b""(bx)/(dt)" "…(1)`
Equation (1) has solution of type
`x=e^(alpha t)`
from (1) `m alpha^(2)+b alpha+k=0`
`alpha=(-b+-sqrt(b^(2)-4mk))/(2m)`
on solving for x, we get
`x=e^((-bt)/(2m))`
`omega_(1)=sqrt(omega_(0)^(2)-alpha^(2))`
where `omega_(0)=sqrt((k)/(m))`
`alpha=(b)/(2)`
`:.` average life `=(2)/(b)`.
So, correct choice is (a).