A particle of mass m is tied at the end of a light string of length L, whose other end is fixed at point C (fig), and is revolving in a horizontal circle of radius r to form a conical pendulum. A parallel horizontal beam of light forms shadow of the particle on a vertical wall.

I
 The time period of the shadow moving on the wall is

(a) The vertical and horizontal component of tension force is equal to mg and m$\mathrm{\omega }$²r respectively ($\mathrm{\omega }$ is angular speed of rotation of the particle). Hence,
$\begin{array}{l}{\mathrm{F}}^2=(\mathrm{mg}{)}^2+{\left({\mathrm{m\omega }}^2\mathrm{r}\right)}^2\\ \Rightarrow {\mathrm{\omega }}^2\mathrm{r}=\frac{1}{\mathrm{m}}\sqrt{{\mathrm{F}}^2-(\mathrm{mg}{)}^2}\\ \mathrm{\omega }={\left(\frac{{\mathrm{F}}^2-(\mathrm{mg}{)}^2}{{\mathrm{m}}^2{\mathrm{r}}^2}\right)}^{1/4}\end{array}$
$\therefore \mathrm{T}=2\mathrm{\pi }{\left(\frac{{\mathrm{m}}^2{\mathrm{r}}^2}{{\mathrm{F}}^2-(\mathrm{mg}{)}^2}\right)}^{\frac{1}{4}}$
This will be the maximum acceleration of the shadow as it will perform SHM with amplitude r and angular frequency $\mathrm{\omega }$.
(b) from the above equation
$\mathrm{\omega }={\left(\frac{{\mathrm{F}}^2-(\mathrm{mg}{)}^2}{{\mathrm{m}}^2{\mathrm{r}}^2}\right)}^{1/4}\therefore \mathrm{T}=2\mathrm{\pi }{\left(\frac{{\mathrm{m}}^2{\mathrm{r}}^2}{{\mathrm{F}}^2-(\mathrm{mg}{)}^2}\right)}^{1/4}$