Question

A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 1.22 s. The temperature rises by 136 C°, and the length of the wire increases. Determine the change in the period of the heated pendulum.

Answer 1

time period of a simple pendulum is given by 2*pi*sqrt(L/g)

where L=length of the pendulum

then 1.22=2*pi*sqrt(L/9.8)

==>L=0.369475 m

coefficient of linear expansion of brass is given by 18.7*10^(-6) per degree celcius

then new length after increase of 136 degree celcius=L*(1+coeffcient*temperature rise)

=0.369475*(1+18.7*10^(-6)*136)=0.370414648 m

then new time period=2*pi*sqrt(length/g)

=1.221549 seconds

then change in time period=0.001549 seconds=1.549 ms

percentage increase=1.549*0.001*100/1.22=0.1297%