Heating a pendulum!

The imbalanced force acting tangent to the pendulum's motion is given by:

$m\ddot{x} \ = \ mg \sin \theta \ \Rightarrow \ \frac{d^2 x}{dt^2} \ = \ g \sin \theta$

Since $x \ = \ R\theta$ , we use the small angle approximation:

$\Rightarrow \ R \frac{d^2 \theta}{dt^2} \ = \ g \sin \theta \ \approx \ g \theta \ \Rightarrow \ \frac{d^2 \theta}{dt^2} \ = \ \frac{g}{R} \ \theta \ \Rightarrow \ \omega \ = \ \frac{2\pi}{T} \ = \ \sqrt{\frac{g}{R}} \ \Rightarrow \ T \ = \ 2\pi \ \sqrt{\frac{R}{g}}$

So the period of the pendulum after the temperature has been increased will be:

$T' \ = \ 2\pi \ \sqrt{\frac{R'}{g}} \ = \ 2\pi \ \sqrt{\frac{R(1 \ + \ \Delta Q\alpha)}{g}}$

So for each full "oscillation" of the pendulum, we lose:

$T' \ - \ T_0 \ = \ 2\pi \ \sqrt{\frac{R(1 \ + \ \Delta Q\alpha)}{g}} \ - \ \ 2\pi \ \sqrt{\frac{R}{g}}$

For a week, measured using "real time" the number of "ticks" of the pendulum will be $n \ = \ \frac{T_{week}}{T_0}$ , thus, the amount of time lost is given by:

$n\Delta T \ = \ \frac{T_{week}}{2\pi \ \sqrt{\frac{R}{g}}} \ \Big( 2\pi \ \sqrt{\frac{R(1 \ + \ \Delta Q\alpha)}{g}} \ - \ \ 2\pi \ \sqrt{\frac{R}{g}} \Big) \ = \ T_{week} \ \Big( \sqrt{1 \ + \ \Delta Q\alpha} \ - \ 1\Big) \ = \ (604 800)(\sqrt{1 \ + \ 1.2 \ \times \ 10^{-4}} \ - \ 1) \ = \ 36.28 \ s$