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? #

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Answer 1 · 2017-12-18T21:38:13

I tried this:

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...

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°alpha = 25 * 10^-6 C°? Is it fast or slow? #