A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 2.15 s. The temperature rises by 133 C°, and the length of the wire increases. Determine the change in the period of the heated pendulum.

Answer 1

T=2πsqrt(L/g) => L=T²g/4π²=
=2.15²•9.8/4•π²=1.1475 m
L₁=L+ΔL = L+αLΔt=
=1.1475 + 19•10⁻⁶•1.1475•133 =1.1504 m
T₁=2 πsqrt(L₁/g) =
= 2 πsqrt(1.1504/9.8)=2.1527 s
ΔT =2.1527-2.15=0.0027 s