Question

A pendulum clock has a pendulum shaft made of aluminum which has a coefficient of linear expansion of 24.0*10^-6K^-1. At 22 degrees celsius the period of the swing of the pendulum is 1.00seconds. If the temperature is reduced is reduced by 0.625°C every hour, how much time does the clock gain or lose in 1 day?

what is its initial length?

Answer 1

T = 2 pi sqrt(L/g)= (2 pi/sqrt g)L^.5 = 1 second at 22 deg so you can get initial L

dT/dL = (2 pi/sqrt g).5 /sqrt L

dT/dt = dT/dL * dL/dt

so what is dL/dt
dL/L = 24*10^-6 dT
dL /dT = 24*10^-6 L
dL/dt = dL/dT *dT/dt
= 24*10^-6 (-.625) L

so
dT/dt = [(2 pi/sqrt g).5 /sqrt L][24*10^-6 (-.625) L ]