A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator a. is going up with an acceleration `a_0`. b. is going down with an acceleration `a_0` and c. is moving with a uniform velocity.

Answer

`=2pisqrt(l/g)`

Explanation

a. Here driving forces
`f=m(g+a_0)sintheta`……….(1)
Acceleration
`a=F/m`
=`(g+a_0)sintheta`
`=(g+a_0)x/l`
(because when q is small `sinthetararr theta=m/l]`
:. `A=((g+a_0)/l)x`
:. Acceleration is proportional to displacement so the motion is SHM.
Now, `omega^2=(g-a_0)/l`
`:. T=2pisqrt(l/((g+a_0))`
b. When the elevator is going downwards with accelerating `a_0`
Driving force
`F=m(g-a_0)sintheta`
Acceleration `=(g-a_0)sintheta((g-a))/lomega^2x`
`T=(2pi)/omega2pi sqrt(l/(g-a_0))`
c. When moving with uniform velocity,
`a_0=0`
For the simple pendulum,
drivig force =`(mgx)/l`
`rarr a=(gx)/l`
`x/a=l/g`
`T=2pisqrt((x)/(a))`
`=2pisqrt(l/g)`