Question

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

Answer 1

Correct Answer - D
`m(d^(2)x)/(dt^(2)) = -kx -b(dx)/(dt)`
`m(d^(2)x)/(dt^(2)) + b(dx)/(dt) + kx = 0` , here `b` is demping coefficient
This has solution of type
`x = e^(lambda_(1))` substituting this
`mlambda^(2) + blambda + k = 0`
`lambda = (-b +- sqrt(b^(2) - 4mk))/(2m)`
on solving for `x`, we get
`x = e^((b)/(2m)t) , a cos (omega_(1) t - alpha)`
`omega_(1) = sqrt(omega_(0)^(2) - lambda^(2))` where `omega_(0) = sqrt((k)/(m))`
`lambda = +(b)/(2)`
So, average life `= (2)/(b)`