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:

Options

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

Answer

D

Explanation

Refer to figure, The retardation of spherical bob due to viscous force `=bupsilon`
`:.` Retarding force `=mb upsilon`
When angular displacemetn of bob is `theta`, the net restoring torque on spherical bob about axis of rotation at `S` is
`tau=-mgxxAP+mb upsilonxxSP`
`=mglsintheta+mbupsilonl` ...(i)
If `alpha` is the angular acceleration of the bob at `P` and `I` is the moment of inertia of the spherical bob about axis of rotation at `S` then
`I=ml^(2)`
and `tau=Ialpha =ml^(2)alpha` ...(ii)
From (i) and (ii)
`ml^(2)alpha=-mglsintheta+mbupsilonl`
or `alpha =-(g)/(l) sin theta +(bupsilon)/(l)`
or `(d^(2)l)/(dt^(2))=-(g)/(l)sintheta+(bupsilon)/(l)`
or `(d^(2)theta)/(dt^(2))+(g)/(l)sin theta-(bupsilon)/(l)=0`
It is a differential equation of second order, involving sine function. Hence it represents dampled oscillations. For small damping the solumtion of the above differential equaltion will be
`theta=theta_(0)e^(-bt//2)sin (omegat+phi)`
Its angular amplitude at time `t=theta_(0)e^(-bt//2)`
As per question ,in time `t=tau` (i.e., average life time ) angular amplitude drops to `(1)/(e)` times its original value `theta_(0)`
`:. (theta_(0))/(e)=theta_(0)e^(-btau//2)` or `e^(-1)=e^(-btau//2)`
or `1=btau//2 or tau =2lb`