A simple pendulum has a time period of 3 s. If the point of suspension of the pendulum starts moving vertically upwards with a velocity =v=kt where `k=4.4 m//s^(2)`, the new time period will be (Take `g=10 m//s^(2))`

Options

A. `(5)/(3)s`
B. 2.5 s
C. 4.4 s
D. `(9)/(4)s`

Answer

B

Explanation

For a simple pendulum,
`T=2pi sqrt((l)/(g))=2pi sqrt((l)/(10))`
But when it is moved upwards, with an acceleratin (a), its effective acceleration becomes g'=g+a.
In this case, `v=kt therefore a=(dv)/(dt)=k=4.4 m//s^(2)`
`therefore g'=g+a=10+4.4=14.4 m//s^(2)`
`therefore T'=2pi sqrt((l)/(g+a))=2pi sqrt((l)/(14.4))`
`therefore (T')/(T)=sqrt((10)/(14.4))=(1)/(1.2)`
`therefore T'=(T)/(1.2)=(3)/(1.2)=(5)/(2)=2.5s`