A 60cm pendulum is fixed in an aeroplane taking off at an angle `alpha=30^(@)` to the horizontal with an acceleration `a=4.0ms^(-2)`. Determine the period of oscillation of the pendulum.

Explanation

Resolving acceleration into two rectangular components, we have
`a_(x)=acos30^(@)=a(sqrt(3))/(2)=0.866a`
`a_(6)=asin3-^(@)=(a)/(2)=0.5a`
The aeroplane flying above can be taken as a llift moving upwards with vertical upward acceleration` a_(y)(=0.5a)`. Therefore, the effective downward acceleration
`a_(y)^(')=g+0.5a`
`:.` Effective acceleration of the pendulum in aeroplane , `g^(')` is
`(g^('))^(2)=(g+0.5a)^(2)+(0.866a)^(2)`
`=(9.8+0.5xx4)^(2)+(0.866xx4)^(2)`
`=151.24`
or `g^(')=12.30m//s^(2)`
The time period of oscillation of the simple pendulum is
`T=2pisqrt((l)/(g^(')))=2pisqrt((0.6)/(12.30))=1.388s`