Question

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? #

Answer 1

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_0alphaDeltaT`

with your data:

`Deltal=1.2*25xx10^-6(8-24)=-0.0005m`

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

the period of the pendulum is:
`T=2pisqrt(l/g)`

at `24^@C`:
`T_1=2pisqrt(1.2/9.8)=2.1986s` each complete oscillation.

at `8^@C`:
`T_2=2pisqrt(1.1995/9.8)=2.1940s` each complete oscillation.

So at `8^@C` the pendulum takes less time (is faster) to complete an oscillation so the clock should be faster...