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

Options

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

Answer

C

Explanation

The equation of SHM,
`(d^(2)x)/(dt^(2))=alphax=0" or "(d^(2)x)/(dt^(2))=-alphax`
Comparing it with the equation of SHM
Comparing it with the equation of SHM
`(d^(2)x)/(dt^(2))=-omega^(2)x,omega^(2)=alpha" or "omega=sqrt(alpha)`
`thereforeT=(2pi)/(omega)=(2pi)/(sqrt(alpha))`.