A simple pendulum of length L and mass m has a spring of force constant k connected to it at a distance h below its point of suspension. Find the frequency of vibrations of the system for small values of amplitude.

Explanation

The frequency of vibration should be greater than that of a simple pendulum since the spring addsan additional restoring force.
`f gt (1)/(2 pi) sqrt((g)/(l))`

We can find the frequency `omega` , which is found in equation for angular SHM : `d^(2) theta//dt^(2) = - omega^(2) theta`. The angular acceleration can be found from analysing the tarque acting on the pendulum.
Fior the pendulum (see sketch)
`Sigma tau = l alpha and d^(2) theta//dt^(2) = - alpha`
The negative sign appears because positive `theta` is measured clockwise in the picture. We take torque around the point of suspension:
`Sigma tau = M g L sin theta + k x h cos theta = l alpha`
For small amplitude vibrations, use the approximations:
`sin theta = cos theta ~~ 1, and x = h tan theta = h theta`
Therefore, with `l = m L^(2)`.
`(d^(2) theta)/(dt^(2)) = - omega^(2) theta`
With angular frequency `omega = sqrt((M gL + kh^(2))/(ML^(2))) = 2 pi f`
The frequency is `f = (omega)/(2 pi) = (1)/(2 pi) sqrt((M gL + kh^(2))/(ML^(2)))`