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? #

1 Answer
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_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...