The frequency of a particle performing a linear S.H.M is `(5)/(2pi)`Hz. The differential equation of S.H.M. is

Options

A. `(d^(2)x)/(dt^(2))+16x=0`
B. `(d^(2)x)/(dt^(2))+25x=0`
C. `(d^(2)x)/(dt^(2))+15x^(2)=0`
D. `(d^(2)x)/(dt^(2))+10x=0`

Answer

B

Explanation

The s"tan"dard equation is `(d^(2)x)/(dt^(2))+omega^(2)x=0`
`omega=2pin=2pi xx(5)/(2pi)=5 " " omega^(2)=25`
`therefore` The equation is `(d^(2)x)/(dt^(2))+25x=0`