If a simple pendulum has significant amplitude (up to a factor of1//e of original) only in the period between `t-0s to t=tau s`, then `tau` may be called the average life of the pendulum. When the sphetical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with b as the constant of propotional to average life time of the pendulum is (assuming damping is small) in seconds:

Explanation

Retarding force `= mbv` When angular displacement of the pendulum is `theta`, net restroing torque `=- mgl sin theta + mbvl` If `I` is the moment of intertial of the pendulum about the point of suspension, then `I alpha = - mgl sin theta + mbvl `Where `alpha = ang. acc.`
Now, as `I = ml^(2), ml^(2) alpha = - mgl sin theta + mbvl`
`:. alpha = (d^(2)theta)/(dt) = - (g)/(l)sin theta +(bv)/(l)`
For small damping, the solution of the above differential equation will be: `theta = theta_(0)e^(-bt//2) sin(omegat + phi)`
`:.` Angular amplitude `= theta_(0)e^(-bt//2)`
According to question, in time `tau` (average life time) angular amplitude reduces to `(1//e)` the value of the orginal value
`:. (theta_(0))/(e) = theta_(0)e^(-bt//2), (bt)/(2) = 1 or t = (2)/(b)`