Determine the natural frequency of vibration of the `100N` disk. Assume the disk does not slip on the inclined surface.

Explanation

In equilibrium, `mg sin theta = kx_(0)`…(i)
When displaced by `x`,
`E = (1)/(2) mv^(2) + (1)/(2) I omega^(2) + (1)/(2)k(x + x_(0))^(2) - mgx sin theta`
Since, E = constant
`(dE)/(dt) = 0`
`0 = mv((dv)/(dt)) + I omega ((d omega)/(dt)) + k (x + x_(0))(dx)/(dt) - mg sin theta (dx)/(dt)`
Substituting, `(dv)/(dt) = a, omega = (v)/(R), I = (1)/(2)mR^(2)`
`(d omega)/(dt) = alpha = (a)/(R),(dx)/(dt) = v`
and `kx_(0) = mg sin theta`
We get, `3ma = - 2kx`
`:. f = (1)/(2x)sqrt|(a)/(x)| = (1)/(2pi) sqrt((2k)/(3m))`
Substituting the values,
`f = (1)/(2pi)sqrt(((2 xx 200)/(3 xx 100))/(9.8))`
`= 0.56 Hz`.