Two spring of force constants `K` and `2K` are connected a mass `m` below The frequency of oscillation the mass is

Options

A. `1/(2pi)sqrt(k/m)`
B. `1/(2pi)sqrt((2k)/(m))`
C. `1/(2pi)sqrt((3k)/(m))`
D. `1/(2pi)sqrt(m/k)`

Answer

C

Explanation

Let the mass m be displaced towards right by a distance x. Then the spring 1 will be extended and the spring 2 will be compressed by an amount of x each. The restoring force on mass m due to spring 1 is `F_1=-kx`
And the restoring force due to spring 2 is
`F_2=-(2k)x`
`therefore` Total force on the mass m is
`F=F_1+F_2=-kx-2kx=-3kx`
`rArr m(d^2x)/(dt^2)=-3kx rArr (d^2x)/(dt^2)+(3k)/(m)x=0`
`rArr (d^2x)/(dt^2)+ omega^2x=0 " where " omega^2 = (3k)/(m) rArr omega = sqrt((3k/(m))`
`rArr = (omega)/(2pi)=1/(2pi)sqrt((3k)/(m))`