If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2))+alphax=0`, its time period is

Options

A. `(2pi)/(alpha)`
B. `(2pi)/(sqrt(alpha))`
C. `2pialpha`
D. `2pisqrt(alpha)`

Answer

B

Explanation

`(d^(2)x)/(dt^(2)) = -alphax………(i)`
We know `a = (d^(2)x)/(dt^(2)) = -omega^(2)x …….(ii)`
From Eq. `(i)` and `(ii)`, we have
`omega^(2) = alpha`
`omega = sqrt(alpha)`
or `(2pi)/(T) = sqrt(alpha) :. T = (2pi)/(sqrt(a))`