A simple pendulum performs simple harmonic motion about `x = 0` with an amplitude a and time period `T` speed of the pendulum at `x = a//2` will be

Options

A. `(pi asqrt(3))/(2T)`
B. `(pi a)/(T) `
C. `(3pi^(2)a)/(T) `
D. `(pi a sqrt(3))/(T)`

Answer

D

Explanation

`v = (dx)/(dt) = A omega cos omega t = A omega sqrt(1 - sin^(2) omegat)`
`= = omega sqrt(A^(2) - y^(2))`
Here `y = (pi)/(2)`
`:. V = omega sqrt(a^(2) - (a^(2))/(4)) = omega sqrt((3a^(2))/(4)) = (2pi)/(T) (asqrt(3))/(2) = (pisqrt(3))/(T)`