# An aluminum pendulum of length 1.2 m keeps accurate time at 24°C .You go on a winter vacation for 14 days. The average temperature of the house is 8°C. How far off will the clock be when you return, if alpha = 25 * 10^-6 C°? Is it fast or slow? #

Dec 18, 2017

I tried this:

#### Explanation:

Cooling the pendolum will reduce its length so that the period will be affected as well.
The length will change by:

$\Delta l = {l}_{0} \alpha \Delta T$

$\Delta l = 1.2 \cdot 25 \times {10}^{-} 6 \left(8 - 24\right) = - 0.0005 m$

so that the new length will be:
$1.2 - 0.0005 = 1.1995 m$

the period of the pendulum is:
$T = 2 \pi \sqrt{\frac{l}{g}}$

at ${24}^{\circ} C$:
${T}_{1} = 2 \pi \sqrt{\frac{1.2}{9.8}} = 2.1986 s$ each complete oscillation.

at ${8}^{\circ} C$:
${T}_{2} = 2 \pi \sqrt{\frac{1.1995}{9.8}} = 2.1940 s$ each complete oscillation.

So at ${8}^{\circ} C$ the pendulum takes less time (is faster) to complete an oscillation so the clock should be faster...