[hamdani yusuf (OP)]

Temperature is a physical quantity that quantitatively expresses the attribute of hotness or coldness. Temperature is measured with a thermometer. It reflects the average kinetic energy of the vibrating and colliding atoms making up a substance.
https://en.m.wikipedia.org/wiki/Temperature

Kinetic energy is 1/2 m.v^2
For oscillating motion, the average velocity might be zero, but the kinetic energy is not.

[Bored chemist]

When a pendulum is at the top of its swing, what is its kinetic energy?
What's the potential energy?
What about when it's at the bottom?
What's the total energy of the swinging pendulum?
Does the total energy change with time (in an idealised case)?
What is the average total energy?

Is the average total energy of the pendulum the same as the average kinetic energy of the pendulum?

[alancalverd]

Kinetic energy is 1/2 m.v^2
For oscillating motion, the average velocity might be zero, but the kinetic energy is not.

Possible misunderstanding here. k.e. is not (m vxv)/2 (a  vector) but (m |v|²)/2 - a scalar multiple of speed, not velocity. The average speed of a pendulum is not zero.

[Bored chemist]

i'm hoping that  hamdani yusuf  has caught on to the fact that the average energy of a vibrating system is NOT the average kinetic energy.
Sorry if I threw in a bit of a red herring earlier. I was making the point that the velocity isn't the only thing happening.
The system still has energy, even when the velocity (or speed) is zero.

[hamdani yusuf (OP)]

When a pendulum is at the top of its swing, what is its kinetic energy?
What's the potential energy?
What about when it's at the bottom?
What's the total energy of the swinging pendulum?
Does the total energy change with time (in an idealised case)?
What is the average total energy?

Is the average total energy of the pendulum the same as the average kinetic energy of the pendulum?

0
Maximum
Maximum, 0
Kinetic+potential
No
=average of Kinetic+potential
No

[hamdani yusuf (OP)]

i'm hoping that  hamdani yusuf  has caught on to the fact that the average energy of a vibrating system is NOT the average kinetic energy.
Sorry if I threw in a bit of a red herring earlier. I was making the point that the velocity isn't the only thing happening.
The system still has energy, even when the velocity (or speed) is zero.

Temperature is more related to kinetic energy instead of the total energy.

[hamdani yusuf (OP)]

Kinetic energy is 1/2 m.v^2
For oscillating motion, the average velocity might be zero, but the kinetic energy is not.

Possible misunderstanding here. k.e. is not (m vxv)/2 (a  vector) but (m |v|²)/2 - a scalar multiple of speed, not velocity. The average speed of a pendulum is not zero.

Where do you get that cross product from?

In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is
1
2
m
v
2
{\textstyle {\frac {1}{2}}mv^{2}}.[2]
https://en.m.wikipedia.org/wiki/Kinetic_energy

In the derivation section, you'll find the use of dot product of vectors, which results in a scalar.

[Bored chemist]

average of Kinetic+potential

... and since  Kinetic+potential is a constant, you don't need to average it.

Temperature is more related to kinetic energy instead of the total energy.

So, a fast moving snow ball (like a comet) is hot...
Well. that's one approach.

[alancalverd]

Where do you get that cross product from?

It represents the square of velocity.

Whilst the average velocity of a pendulum is zero, its average speed is not.

[hamdani yusuf (OP)]

... and since  Kinetic+potential is a constant, you don't need to average it.

Only in ideal case.

[hamdani yusuf (OP)]

So, a fast moving snow ball (like a comet) is hot...
Well. that's one approach.

You seem to forget about entropy.

[alancalverd]

Only in ideal case.

Physics is about ideal gases, weightless strings, etc. So you can assume your pendulum is swinging in a vacuum, suspended from a frictionless pivot.

Engineering is about the real world and uses lots of physics to approximate frictional losses, viscous drag, etc. But the conservation of energy is always observed.

[hamdani yusuf (OP)]

Where do you get that cross product from?

It represents the square of velocity.

Whilst the average velocity of a pendulum is zero, its average speed is not.

In physics, the symbol x is typically used for vector cross product, which is not what you were trying to describe.

The square of a vector typically refers to the square of its magnitude (or length), not the vector itself. Mathematically, if v is a vector with components , the square of its magnitude is given by:

|\mathbf{v}|^2 = v_1^2 + v_2^2 + \dots + v_n^2

This is essentially the dot product of the vector with itself:

\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2

So, squaring a vector usually means taking the sum of the squares of its components, yielding a scalar value (not another vector).

ChatGPT

[Bored chemist]

... and since  Kinetic+potential is a constant, you don't need to average it.

Only in ideal case.

The "ideal" case where energy is conserved.
LOL

[hamdani yusuf (OP)]

Physics is about ideal gases, weightless strings, etc. So you can assume your pendulum is swinging in a vacuum, suspended from a frictionless pivot.

It depends on the level of complexity. High school physics already takes friction into account.

[alancalverd]

In physics, the symbol x is typically used for vector cross product, which is not what you were trying to describe.

But it has the property that you wanted, i.e. an average of zero for harmonic motion.

[Bored chemist]

Physics is about ideal gases, weightless strings, etc. So you can assume your pendulum is swinging in a vacuum, suspended from a frictionless pivot.

It depends on the level of complexity. High school physics already takes friction into account.

Do you think that the zero point vibration of a lattice has friction?
Do you know what an analogy is?

[hamdani yusuf (OP)]

Do you know what an analogy is?

a comparison between two things, typically for the purpose of explanation or clarification.
a process of arguing from similarity in known respects to similarity in other respects.

[Bored chemist]

OK, so, do you recognise that the pendulum was an analogy for the ZPE?
And do you realise that, in this context, there's no energy loss?

[hamdani yusuf (OP)]

OK, so, do you recognise that the pendulum was an analogy for the ZPE?
And do you realise that, in this context, there's no energy loss?

Where did you make that statement?

My previous comment was a response to Alan's statement on the scope of physics.