Period of pendulum with attached mass at distance x from pivot

[nathangrand]

A light rigid pendulum of length l with a bob of mass M is to be used as the
timing element of a mechanical clock. Write down the equation of motion for the
system and hence determine the period of oscillation T. A second mass m is now attached to the pendulum arm at a distance x from the pivot, where 0 < x < l. Obtain an expression for the new period of oscillation T0. Assuming that the mass m is much smaller than M, determine the position
x at which it exerts its greatest infuence on the period T0.

For first part got d''(theta)/dt^2 (g/l)theta =0
which gives T=2Pi*SQRT(L/g)

For second part get equation d''(theta)/dt^2 + theta(Mgl + mgx)/(mx^2 + Ml^2) = 0
and T0 of 2pi* SQRT((mx^2 + Ml^2)/(Mgl + mgx))

I have no real idea though how to find the value for x at which the small mass has the largest influence on T0...thought about differentiating something perhaps?

 

[tiny-tim]

hi nathangrand! welcome to pf!

(have a square-root: √ and a theta: θ and a pi: π and try using the X² icon just above the Reply box )

For second part get equation d''(theta)/dt^2 + theta(Mgl + mgx)/(mx^2 + Ml^2) = 0
and T0 of 2pi* SQRT((mx^2 + Ml^2)/(Mgl + mgx))

you want to find the greatest effect of that square-root, of the form √(Ax² + B)/(Cx + D) …

so forget the square-root, and deal with (Ax² + B)/(Cx + D) on its own

 

[nathangrand]

Ok..I have done this and differentiated it with respect to x to get :

mx^2 + 2MLx - ML^2 = 0

with a bit of messing around with the quadratic formula I get x=L/2 is this right??

 

[tiny-tim]

hi nathangrand!

without seeing your full calculations, i can't check,

but L/2 is certainly the answer i'd have guessed!