The work done by the tension in the strings of a simple pendulum in one complete oscillation is equal to

Explanation

Consider a conical pendulum of string length L with its bob of mass m performing UCM along a path of radius r. Let v be the cosntant linear speed of the bob and `vecF` the tension in the string.
Resolve `vecF` into components `F cos theta` vertically opposite to the weight of the bob and `F sin theta` horizontal, `F cos theta` balances the weight . `F sin theta` is teh necessary centripetal force.
`therefore F sin theta = (mv^(2))/(r )" '...(1)`
and `F cos theta = mg " "..(2)`
Dividing Eq. (1) by Eq. (2),
`tan theta = (v^(2))/(rg) " "...(3)`
Also, from the diagram,
`tan theta = (PC)/(OC) (r)/(h)" "...(4)`
where OC = h = L cos `theta` is the axial height of the cone. Therefore, from Eqs. (3) and (4),
`(v^(2))/(rg) = (r )/(h) " "therefore (v^(2))/(r ) = (rg)/(h) " "...(5)`

Squaring and adding Eqs. (1) and (2)
`F^(2)sin^(2) theta + F^(2) cos^(2) theta = ((mv^(2))/(r))^(2)+(mg)^(2)`
`therefore F^(2)=((mrg)/(h))^(2)+(mg)^(2)=(mg)^(2)[((r)/(h))^(2)+1]`
`therefore F = mgsqrt(((r)/(h))^(2)+1)=mg sqrt(((r)/(Lcostheta))^(2)+1)`
This is the expression for the tension in the string.