Question

A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 1.17 s. The temperature rises by 140 C°, and the length of the wire increases. Determine the change in the period of the heated pendulum.

Answer 1

here,

let the initial length of wire be l0

time period , T0 = 2 *pi * sqrt(l0/g)

1.17 = 2 *pi * sqrt(l0/9.81)

l0 = 0.3405 m

linear expansion coefficient of brass , l = 19 * 10^-6 per degree C

change in temprature , dT = 140 degree C

the new length , l = l0 * ( 1 + alpha * dT)

l = 0.3405 * ( 1 + 140 * 19 * 10^-6)

l = 0.3414 m

the new time period , T = 2*pi* sqrt(l/g)

T = 2 *pi * sqrt(0.3414/9.81)

T = 1.17154 s

the change in period , dT = T - T0 = 1.55 * 10^-3 s