Question

A pendulum clock of time period 2 sec gives the correct time at ${30}^{\circ }C$. The pendulum is made of iron. How many seconds will it lose or gain per day when temperature falls to $0^{\circ }C$ ?
${\alpha }_{Fe}=1.2\times {10}^{-5}{(}^{\circ }C{)}^{-1}$

Answer 1

New time period will be given by

$T^{\prime }=2\pi \sqrt{\frac{l(1+\alpha \Delta T)}{g}}=2\sqrt{1+\alpha \Delta T}$

$T^{\prime }=2\sqrt{1+(1.2\times {10}^{-5})(-30)}\approx 2(1-18\times {10}^{-5})s$

Thus, time gained in a day is $24\times 3600\times 36\times {10}^{-5}s=15.55s$.