Question

An aluminum clock pendulum having a period of 1.0000 s keeps perfect time at 20.0°C. When placed in a room at a temperature of -10.2°C, will the clock gain time (run fast) or lose time (run slow)? use 24 x 10^-6 for the expansion coefficient of aluminum.

Answer 1

Answer:

The clock will run slower

Explanation:

Hi!

The period T of the clock pendulum is related to the length of the wire l

$T = 2 \pi  \sqrt{\frac{l}{g}}$

However, when the temperature decreases, the length of the wire will decrease, and therefore the period will decrease.

How much you may ask

We can figure this out dividing the period T at room temperature and T' at -10.2 degrees:

$\frac{T}{T'}= \sqrt{\frac{l}{l'}}$

and using the formula for linear thermal expansion:

$l' = l (\alpha  \text{ \Delta T}+1)$

Therefore:

$\frac{T}{T'}= \sqrt{\frac{1}{(\alpha  \text{ \Delta T}+1)}}$

Where

     α=25x10^-6    1/°C

    ΔT = -10.2 -20 = -30.2°C

Therefore

$\frac{T}{T'}=1.00038$

Or

T' = 0.999622 T

It will run at a 99.96% of its original speed