At what depth is the phase of temperature profile reversed?

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The amplitude of daily temperature variation is 5°C, and the ground is composed of limestone. The depth at which the temperature profile reverses, resulting in cooler temperatures at 3 PM and warmer temperatures at 3 AM, is calculated to be approximately 7.9 meters. This reversal occurs where the cosine function changes sign, specifically at a phase of pi. To determine the ratio of temperature variation amplitude at 7.9 meters compared to the surface, one must consider the thermal conductivity of the limestone, as the depth of temperature variation is influenced by this property. The circular frequency is noted as 2E-7 per second, but further calculations indicate that the angular frequency should be approximately 72.722 x 10^-6 rad/sec based on a 24-hour period. This scenario is modeled by a one-dimensional thermal diffusion equation.
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The amplitude of daily temperature is dT=5 C. The ground is made of limestone.

At what depth is the phase of temperature profile reversed (cooler at 3 pm and warmer at 3 am)?

Circular frequency is equal to I think 2E-7 per second

The depth at which the profile is reversed is where the cosine function switches signs, i.e. at pi.

So, I find that the profile is reversed at 7.9 meters. Does this seem correct?

How could I find the ratio of the amplitude of temperature variation at 7.9 m vs. the temperature variation at the surface?
 
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Replying so that this thread isn't a no-reply, as per Greg's wishes.
If anyone sees this and has knowledge on the subject, feel free to respond.
 
rmjmu507 said:
Circular frequency is equal to I think 2E-7 per second
I would question that.
Period T = 24 hours = 24 * 60 * 60 = 86400 sec .
Frequency in Hz = 1 / T ;
Angular freq = 2 * Pi / 86400 = 72.722x10-6 rad/sec.

This problem is a one dimensional thermal diffusion equation.
The depth will depend on the thermal conductivity of the limestone.
 
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