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I'm trying to solve this task:

The pendulum clock was transported from the Earth's equator to Antarctica (in the vicinity of the southern geopole) for scientific experiments. Estimate the pendulum clock correction over the Earth's day in Antarctica (at a temperature of $t = −90 ° C$) if these clocks are calibrated at the equator (at a temperature of $t = + 50 ° C$). The coefficient of thermal expansion of the pendulum substance is $α_h$ = $2,4 · 10^−5$ $deg^−1$. The original verified length of the pendulum is $ℓ_0$ = 300 mm. How much should the length of the pendulum be changed so that the correction of the hours per day is no more than 10 seconds?

I think I have almost solved the task, but at the very last moment a have my variable being gone... Here is the solution:

Formulas to use: $L_t=L_0 × (1+α*t)$, $T=2π×√(l/g)$

We should find $L_0$ first:

$L_0=L_{+50}/(1 + 2,4 × 10^{-5} × 50)=0,3/(1+0,0012)=0,29964 (m)$

Then $L_{-90}$:

$L_{-90}=L_0×(1+2,4×10^{-5}×(-90))=0,29964×(1-0,00216)=0,29964×0,99784=0,29899 (m)$

Periods:

$T_{+50}=2π × √(L_{+50}/g)=2 × 3,142 × √(0,3/9,81)=1,0989 (sec)$

$T_{-90}=2π × √(L_{-90}/g)=2 × 3,142 × √(0,29899/9,81)=1,09706 (sec)$

$∆T=T_{+50}-T_{-90}$

After that we should find the number of complete fluctuations per day:

$N=(24 × 60 × 60)/T_{-90} =86400/1,09706=78755,9477$

So the correction should be:

$x=N × ∆T=78755,9477 × (1,0989-1,09706)=144,91 (sec)$

Now we need it need to be less than 10 secs:

$N × ∆T<10$

Here just transformations:

$\frac{86400}{(2π × √(L_{-90}/g))}×(2π× √(L_{+50}/g)-2π× √(L_{-90}/g))<10$

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And so the $L_{+50}$ just disappears! I must have made a mistake somewhere. Could somebody help?

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    $\begingroup$ $g$ varies quite a bit from the equator to antarctica. One slightly annoying thing is your use of $*$ to represent multiplication. Either $\times$ or $\cdot$ or nothing at all is prefered $\endgroup$
    – James K
    Nov 28 '21 at 10:26
  • $\begingroup$ Thanks for the remind of $g$, I also edited my question with proper multiplication signs. $\endgroup$
    – ALiCe P.
    Nov 28 '21 at 10:30
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There are a couple of oddities I see in this working.

I'm not completely clear if $l_0$ is the length at zero celcius or at the equator, at 50 celcius. But I'll go with your interpretation, that 300 = length at 50 celcius.

The temperatures seem extreme and unreasonable. It is warm at the equator and cold in Antarctia, but +50 to -90 seems rather unreal!

Given that, the next thing is your use of 9.81 for g. This is simply not correct at the equator or at the pole. At the equator you should use 9.78, and at antarctica you need 9.83

That said, the final part is just to find the proper length in antarctica, using the local value of g. That will be slightly longer than 300mm due to the higher gravitational acceleration.

You would just write down the formula for the period and set it equal to the time period at the equator, and use that to find the length. The difference in length is the change you need to make. You can then investigate how errors in the value of "g" used propagate and if this allows for the required accuracy.

If necessary you can convert that length to a length at the pole. (and its not clear to me if the change in length needs to be done at the pole or at the equator) Again there is a question of error propagation.

I think this question is really about the variation in $g$. If $g$ is assumed to be constant, then the length of the pendulum at the pole should be 300mm to get the same period as at the equator, since the period of a pendulum depends only on the length and on the local gravitational field. And that doesn't seem to be an interesting question. At least if $g$ is assumed constant then this question is off topic, as there is no astronomical content at all!

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(1.1)^2>= L(-90)/L(+50) >= (0.9)^2 L(-90) = L(+50) *(1+a.dT)=> x.Lo(1+a.dT) >= 0.81 Lo and L(-90) = L(+50) *(1+a.dT)=> x.Lo(1+a.dT) <= 1.21 Lo

Gives us x >=0.81/(1+a.dT) and x<=1.21/(1+a.dT) where x is the ratio of L/Lo and 1-x is the % change in length So yes, Lo should disappear

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