1.

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 :

Answer»

`(2)/(B)`
`(0.693)/(b)`
b
`(1)/(b)`

Solution :`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).


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