Question

Figure. Shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is

Answer 1

The slope of the graph
`=(F)/(x) = (-0.5)/(5) = 0.1N cm^(-1) = -10 N m^(-1)`, But `F =- m omega^(2)x`
or `(F)/(x) =- m omega^(2) so -m omega^(2) =- 10 or m omega^(2) = 10` or
`omega^(2) =(10)/(m), :. Omega^(2) = (10)/(4xx10^(-1)) rArr omega =(10)/(2) = 5, :. f = (omega)/(2pi) = (5)/(2pi) s^(-1)`