How can I accurately calculate the quantum number of a pendulum?

[jjson775]

Summary:: Calculate the quantum number of a pendulum

I want to calculate the quantum number of pendulum. L = 1m, m = 1 kg., A= 3cm. I get a period of 2.01 sec. and f = 1/T = .498 sec. E =nhf gives me 2.67x10^31. The correct answer is 1.33x10^31, Where am I going wrong?

[Thread moved from the technical forums, so no Homework Template is shown]

 

[PeroK]

Summary:: Calculate the quantum number of a pendulum

I want to calculate the quantum number of pendulum. L = 1m, m = 1 kg., A= 3cm. I get a period of 2.01 sec. and f = 1/T = .498 sec. E =nhf gives me 2.67x10^31. The correct answer is 1.33x10^31, Where am I going wrong?

It's probably this again:

https://www.physicsforums.com/threads/de-broglie-relations-confusion.987026/#post-6324041

 

[etotheipi]

We have $E = T+V = mgl(1-\cos{0.03})$ and $f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$, along with $E = nhf$ for the oscillator.

With that I get the same as the book answer. How did you calculate the energy?

 

[Vanadium 50]

Where am I going wrong?

If you want us to answer that, shouldn't you show your work?

 

[etotheipi]

Here's a question that I'll put out there in case anyone has an answer (I don't intend for it to detract from the OP, so if it is a digression then let me know and I'll delete it!).

$V$ can be shifted by an additive constant, so $E = nhf$ could equal any real number. What does that mean for our calculated quantum number?

 

[PeroK]

Here's a question that I'll put out there in case anyone has an answer (I don't intend for it to detract from the OP, so if it is a digression then let me know and I'll delete it!).

$V$ can be shifted by an additive constant, so $E = nhf$ could equal any real number. What does that mean for our calculated quantum number?

That would increase the ground state energy, and all energy levels. It won't give a different $n$.

 

[etotheipi]

That would increase the ground state energy, and all energy levels. It won't give a different $n$.

Cool, thanks.

 

[PeroK]

Summary:: Calculate the quantum number of a pendulum

I want to calculate the quantum number of pendulum. L = 1m, m = 1 kg., A= 3cm. I get a period of 2.01 sec. and f = 1/T = .498 sec. E =nhf gives me 2.67x10^31. The correct answer is 1.33x10^31, Where am I going wrong?

[Thread moved from the technical forums, so no Homework Template is shown]

The problem is that $T$ and $f$ are constant for a given $L$, hence the energy level of the pendulum is independent of $f$. It's interesting that you got exactly twice the correct answer. It's not what I suggested in post #2.

 

[jjson775]

We have $E = T+V = mgl(1-\cos{0.03})$ and $f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$, along with $E = nhf$ for the oscillator.

With that I get the same as the book answer. How did you calculate the energy?

I assumed the energy to be the potential energy at maximum displacement, ie, mgy =

We have $E = T+V = mgl(1-\cos{0.03})$ and $f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}$, along with $E = nhf$ for the oscillator.

With that I get the same as the book answer. How did you calculate the energy?

Thanks. I got the potential energy wrong and fixed it. I am a retired engineer who studied classical physics only, 60 years ago, and am very interested in modern physics, at least at the undergraduate level for engineering students. I got through special relativity and am starting quantum mechanics. This is my first post on this forum.

 

[PeroK]

I assumed the energy to be the potential energy at maximum displacement, ie, mgy =

Thanks. I got the potential energy wrong and fixed it. I am a retired engineer who studied classical physics only, 60 years ago, and am very interested in modern physics, at least at the undergraduate level for engineering students. I got through special relativity and am starting quantum mechanics. This is my first post on this forum.

Which book are you working from? It's only my view, but this question isn't really about QM. A $1kg$ pendulum is a macroscopic, decoherent multi-particle system which does not in any meaningful way have a quantum number.

I'm not sure what you learn about QM from a question like this.

 

[kuruman]

Which book are you working from? It's only my view, but this question isn't really about QM. A $1kg$ pendulum is a macroscopic, decoherent multi-particle system which does not in any meaningful way have a quantum number.

I'm not sure what you learn about QM from a question like this.

Here, this decoherent multi-particle system is no more unphysical than a point car passing a point truck on the highway in a kinematics problem. My view is that we have to suspend our disbelief in order to abstract the main point to be made. In the car problem, the point is to illustrate the application of the kinematic equations to a "catch-up" situation. In this example the point is to illustrate how the Correspondence Principle works. When $n$ is of order $10^{31}$, one can better understand the classical limit of "continuous" energy by looking at the energy difference between the $n$th and $(n+1)$th energy levels of the oscillator.

 

[PeroK]

Here, this decoherent multi-particle system is no more unphysical than a point car passing a point truck on the highway in a kinematics problem. My view is that we have to suspend our disbelief in order to abstract the main point to be made. In the car problem, the point is to illustrate the application of the kinematic equations to a "catch-up" situation. In this example the point is to illustrate how the Correspondence Principle works. When $n$ is of order $10^{31}$, one can better understand the classical limit of "continuous" energy by looking at the energy difference between the $n$th and $(n+1)$th energy levels of the oscillator.

If we calculate the stopping distance of a car, modelling it as a point particle, then that has some physical meaning. If I say a car traveling at 100km/h has a stoping distance of 50m, that has some meaning.

If you say a pendulum is in the energy state $10^{31}$ it's meaningless, especially as the pendulum does not meet the basic criteria to be modeled as a quantum SHO.

 

[jjson775]

Which book are you working from? It's only my view, but this question isn't really about QM. A $1kg$ pendulum is a macroscopic, decoherent multi-particle system which does not in any meaningful way have a quantum number.

I'm not sure what you learn about QM from a question like this.

Serway and Beichner Physics for Scientists and Engineers. It is an older edition but suits my purposes. I believe it is very widely used. There is a similar sample problem in the book that shows calculation of the quantum number for an oscillating spring block system. The book says that even though the energy is quantized, the effect is only measurable or important on the level of atoms and molecules.

 

[kuruman]

If we calculate the stopping distance of a car, modelling it as a point particle, then that has some physical meaning. If I say a car traveling at 100km/h has a stoping distance of 50m, that has some meaning.

If you say a pendulum is in the energy state $10^{31}$ it's meaningless, especially as the pendulum does not meet the basic criteria to be modeled as a quantum SHO.

It looks like you misunderstood my points on both accounts. With the car and truck problem there was no question about stopping distance. The stopping distance is the same for all points on a decelerating car assuming that it is a rigid body. When the question is "when does the car pass the truck?", one has to model each vehicle as a point object otherwise there will be a non-zero time interval $\Delta t$ during which a point on the car is at the same position with a point on the truck in which case the question would become meaningless unless on specifies the length of each vehicle and has defined what "passing the truck" exactly means. However, that would detract from the pedagogical value of the question which is the application of the kinematic equations.

Likewise for the 1 kg pendulum as a quantum harmonic oscillator. In all textbook descriptions of the quantum oscillator that I have seen, there are no restrictions or footnotes specifying the circumstances under which it can be applied. Specifically, I have seen no criteria specifying the mass $m$ in the Hamiltonian and when the wavefunctions are derived, there are no criteria about what values of $n$ are appropriate. I have always viewed the harmonic oscillator and its companion the particle in a box as two of the few examples in QM that can be solved exactly and can be used to show the transition from the quantum to the classical regime. Yes, $n=10^{31}$ is a ridiculously meaningless number, but that's not the point. The point is that the level with $n=10^{31}+1$ is ridiculously close to the level with $n$ so one might as well call that a continuum of energy, which is what we've been doing before QM.

Anyway, these are my thoughts and we can agree to disagree.

 

[PeroK]

Likewise for the 1 kg pendulum as a quantum harmonic oscillator. In all textbook descriptions of the quantum oscillator that I have seen, there are no restrictions or footnotes specifying the circumstances under which it can be applied. Specifically, I have seen no criteria specifying the mass $m$ in the Hamiltonian and when the wavefunctions are derived, there are no criteria about what values of $n$ are appropriate. I have always viewed the harmonic oscillator and its companion the particle in a box as two of the few examples in QM that can be solved exactly and can be used to show the transition from the quantum to the classical regime. Yes, $n=10^{31}$ is a ridiculously meaningless number, but that's not the point. The point is that the level with $n=10^{31}+1$ is ridiculously close to the level with $n$ so one might as well call that a continuum of energy, which is what we've been doing before QM.

Anyway, these are my thoughts and we can agree to disagree.

What would be valid is starting with QM to derive the equations of motion of a macroscopic pendulum. But, what is done here is to use Newton's laws of motion to derive the equation of motion of a pendulum and then ascribe a quantum number to the solution. This is fundamentally invalid, IMHO. If the pendulum is a quantum object it obeys the SDE and not Newton's laws.

So, first things first: please put a particle of mass $m$ in a quantum harmonic oscillator and derive the classical equations of motion!

 

[kuruman]

What would be valid is starting with QM to derive the equations of motion of a macroscopic pendulum. But, what is done here is to use Newton's laws of motion to derive the equation of motion of a pendulum and then ascribe a quantum number to the solution. This is fundamentally invalid, IMHO. If the pendulum is a quantum object it obeys the SDE and not Newton's laws.

In this specific example, OP deduced the frequency and energy of the motion from the length of the pendulum, the amplitude and the value of g. Doing that admittedly used equations derived from Newton's laws. Is that what you object to? What if the question simply gave the energy and frequency of the macroscopic pendulum and asked for the quantum number $n$? Couldn't one use $E_n=(n+\frac{1}{2})\hbar \omega$ which is the QM-derived result?

 

[PeroK]

In this specific example, OP deduced the frequency and energy of the motion from the length of the pendulum, the amplitude and the value of g. Doing that admittedly used equations derived from Newton's laws. Is that what you object to? What if the question simply gave the energy and frequency of the macroscopic pendulum and asked for the quantum number $n$? Couldn't one use $E_n=(n+\frac{1}{2})\hbar \omega$ which is the QM-derived result?

My point is that the pendulum is not a quantum mechanical SHO, it's a classical SHO (which may be describable as a complex QM system, but not as QM SHO). It's not, therefore, in a well-defined QM SHO energy eigenstate of whatever $n$ one can calculate.

If one could produce an elementary particle with a mass of $1kg$, that might be interesting: that, I assume, would be smeared out without a well-defined trajectory and behave quantum mechanically. But, a $1kg$ object comprising trillions of elementary particles does not behave like a hypothetical elementary particle of $1kg$.

 

[kuruman]

I get your point. I am trying to find some educational value in this while your position is that there can be none.