A body is oscillating vertical about its mean position with period `sqrt(3)` s. If its maximum and minimum heights above the surface of the ground are 2 m and 1.5 m, then the speed at midway between mean and extreme position will be

Options

A. `5 m//s`
B. `(pi)/(4) m//s`
C. `10 m//s`
D. `(pi)/(sqrt(3)) m//s`

Answer

B

Explanation

`2A=2-1.5 therefore A=(0.5)/(2)=0.25m`
`v=omegasqrt(A^(2)-x^(2))therefore x=A//2`
`v=(sqrt(3))/(2)Aomega=(sqrt(3))/(2)xx(1)/(4)xx(2pi)/(sqrt(3))`
`=0.866 A omega`