A compensated pendulum shown in figure is in the form of an isosceles triangle of base length l_(1)=10cm and coefficient of linear expansion alpha_(1)=18xx10^(-6) and side length l_(2) and coefficient of linear expansion alpha_(2)=12xx10^(-6). Find l_(2) so that the distance of centreof mass of the bob-from suspension centre O may remain the same at all the temperatures.

A compensated pendulum shown in figure is in the form of an isosceles

1.	A compensated pendulum shown in figure is in the form of an isosceles triangle of base length l_(1)=10cm and coefficient of linear expansion alpha_(1)=18xx10^(-6) and side length l_(2) and coefficient of linear expansion alpha_(2)=12xx10^(-6). Find l_(2) so that the distance of centreof mass of the bob-from suspension centre O may remain the same at all the temperatures.
Answer» SOLUTION : In this case the distance of center of mass from suspensionpoint O is `h=sqrt(l_(2)^(2) -(l_(1)^(2))/(4))` At some higher temperature `Deltat`, new value of h will become `h.=sqrt(l_(2)^(2)(1+alpha_(2) Deltat)^(2) -(l_(1)^(2))/(4)(1+alpha_(1)Deltat)^(2)) ` As we require `h=h.`, we have `l_(2)^(2)-(l_(1)^(2))/(4)=l_(2)^(2)(1+2alpha_(2) Deltat)-(l_(1)^(2))/(4)(1+2alpha_(1)Deltat)` (or) `alpha_(2)l_(2)^(2) =(l_(1)^(2))/(4) alpha_(1) or l_(2) =(l_(1))/(2) sqrt((alpha_(1))/(alpha_(2))) =(10)/(2) sqrt((18xx10^(-6))/(12xx10^(-6)))=6.12cm`