Harmonic oscillator and simple pendulum time period

[Misha87]

TL;DR

Comparing a simple harmonic oscillators' and a simple pendulums' period T

Hi, I have been thinking about pendulums a bit and discovered that a HO(harmonic Oscillator) will take the same time to complete one period T no matter which amplitude A/length l it has, if stiffness k and mass m are the same.
But moving on to a simple pendulum suddenly the time period for one oscillation changes with its length l.
For a simple pendulum mass is negligible and can therefore not be considered a falling body? Is that the reason? How would the formula look if the pendulum was a simple HO?

 

[malawi_glenn]

For the pendulum, with small angle approximation $\sin \theta \approx \theta$ the period is $T = 2 \pi \sqrt{\dfrac{\ell}{g}}$

The mass is not negligible, it just turns out that it cancels out when you write down the equation of motion (see below).

For an harmonic oscillator, i.e. ideal spring with spring consant $k$ the period is $T = 2 \pi \sqrt{\dfrac{m}{k}}$ where $m$ is the mass.

You can very easily see that $\dfrac{\ell}{g} = \dfrac{m}{k}$ thus the length of the pendulum play the same rôle as mg/k in the spring HO system.

Remember that the corresponding amplitude (well, deviation from equilibrium) in the pendulum is the angle $\theta .$ This means that regardless of your $\ell$ and $g$, as long as the small approximation is valid, the period T is independent of the amplitude.

The equation of motion for the pendulum (small angle approximation) is $m \ell^2 \ddot \theta + mg\ell \theta = 0$ which is just the formula: sum of external torques = moment of inertia times angular acceleration. Thus, displacements from the equilibrium in the pendulum is measured in angle (radians).

The equation of motion for the HO system is $m \ddot x + k x = 0$ which is just Newtons second law: sum of external forces = mass times acceleration. Thus, displacements from the equilibrium in the pendulum is measured in linear distance (meters).

To conclude, here is a "lexicon"

Pendulum	HO
$\ell$	$mg/k$
$\theta$	$x$

I hope this helped a bit.

 

[Misha87]

Thank you for your quick response!
I do see both formulas but they don't make sense to me.
I am thinking of a grandfather clock, for example. I was thinking that the pendulum should take the same time T to complete one complete oscillation, no matter which length l the pendulum has. If the pendulum was shortened then the pendulum will swing faster from one max to another and if it is longer then it will swing slower (amplitudes A stays the same in both cases). This is how I understand the HO.
But, the formula of the simple pendulum implies that T changes if the length is changed. This does not make sense to me...

 

[malawi_glenn]

Note that the period for the pendulum is independent of the angle $\theta$ just as the period for the HO is independent of the linear displacement $x$. The angle $\theta$ is independent of $\ell$, this you must realize.

Thus, the periods are independent of the amplitude. (as long as you can use the small angle appriximation for the pendulum)

Changing $\ell$ means basically changing $mg/k$ in the HO. The longer pendulum you have, the longer period T.

But, for a fixed $\ell$, you can release the pendulum from any starting angle (amplitude) and you get the same period, again as long as you can use the small angle appriximation for the pendulum.

I was thinking that the pendulum should take the same time T to complete one complete oscillation, no matter which length l the pendulum has

this is wrong thinking. This is why math is needed in physics. Our "intuition" is not good enough.

You can actually do a simple unit analysis, you need the unit s on one side. Now what variables do you have to play with? You have length m and you have acceleration due to gravity m/s². There is no way you can end up with units of s on one side (period) without invoking some length on the other side:
$[T] = [\ell]^a [g]^b$
$\text{s} = \text{m}^a \text{m}^b \text{s}^{-2b}$
solution: $b = -1/2$ and $a = 1/2$
which results in
$T = \text{constant}\cdot \sqrt{\ell / g}$

I love to fool my students on tests with questions like this:
An HO is released at rest 4,0 cm from equilibrium and undergoes harmonic motion with T = 0,25 s.
What would the period be if it instead was relased at rest 8,0 cm from equilibrium?

Not that many correct answers, in fact, even the most brilliant students misses out on these kind of questions.

 

[vanhees71]

The simple pendulum has another equation of motion than a harmonic oscillator, i.e., the driving force is not proportional to the displacement from the equilibrium position:
$$\ddot{\vartheta}=-\frac{g}{l} \sin \vartheta.$$
That's why the period depends on the amplitude.

It was an ingenious finding by Huygens that for independence of the period of the amplitude you have to make the mass move along a cycloide rather than a circle, and he constructed a clock based on this finding by making use of the fact that the evolute of a cycloid is itself again a cycloid:

https://en.wikipedia.org/wiki/Cycloid#Cycloidal_pendulum

 

[malawi_glenn]

The simple pendulum has another equation of motion than a harmonic oscillator, i.e., the driving force is not proportional to the displacement from the equilibrium position:
$$\ddot{\vartheta}=-\frac{g}{l} \sin \vartheta.$$
That's why the period depends on the amplitude.

It was an ingenious finding by Huygens that for independence of the period of the amplitude you have to make the mass move along a cycloide rather than a circle, and he constructed a clock based on this finding by making use of the fact that the evolute of a cycloid is itself again a cycloid:

https://en.wikipedia.org/wiki/Cycloid#Cycloidal_pendulum

Yeah if the small angle approximation is no longer valid, the period of the pendulum indeed depends on the amplitude :)

 

[Misha87]

this is wrong thinking. This is why math is needed in physics. Our intuition is not good enough.

Thanks for clearing this up for me, it makes sense now, that I realize I had a mistake in my initial thought experiment.

 

[anorlunda]

am thinking of a grandfather clock, for example. I was thinking that the pendulum should take the same time T to complete one complete oscillation, no matter which lengt

Test out yourself. All pendulum clocks have an adjustment nut at the bottom of the pendulum. If the clock is to fast or to slow, turn the nut.

Simpler still just swing a mass on a string try it with 10 cm length, then 100 cm length.

 

[malawi_glenn]

Thanks for clearing this up for me, it makes sense now, that I realize I had a mistake in my initial thought experiment.

Note however, as was pointed out, when you no longer can assume the small angle approximation, you will have that the amplitud of the pendulum enters the period.
$T = 2 \pi \sqrt{\ell / g} \left( 1 + \theta_\text{max}^2/16 + 11\theta_\text{max}^4/3072 + \ldots \right)$
For $\theta_\text{max} = 30^\circ$ you are off by approximately 2% from the small angle approximaton period.
For $\theta_\text{max} = 60^\circ$ you are off by approximately 7% from the small angle approximaton period.
You might not think that 2% is that much, but in 24h it is about 30 min so. Now there are other effects you could worry about, like friction etc.

 

[bob012345]

Note however, as was pointed out, when you no longer can assume the small angle approximation, you will have that the amplitud of the pendulum enters the period.
$T = 2 \pi \sqrt{\ell / g} \left( 1 + \theta_\text{max}^2/16 + 11\theta_\text{max}^4/3072 + \ldots \right)$
For $\theta_\text{max} = 30^\circ$ you are off by approximately 2% from the small angle approximation period.
For $\theta_\text{max} = 60^\circ$ you are off by approximately 7% from the small angle approximation period.
You might not think that 2% is that much, but in 24h it is about 30 min so. Now there are other effects you could worry about, like friction etc.

What is the typical amplitude of a grandfather clock?

 

[malawi_glenn]

What is the typical amplitude of a grandfather clock?

idk 5 degrees?

Grandfather clocks also have a mechanism for "brining back" energy into the system (anchor escapement). So one can not treat a grandfather pendulum as a simple pendulum really.

 

[bob012345]

idk 10 degrees?
Grandfather clocks also have a mechanism for "brining back" energy into the system (anchor escapement). So one can not treat a grandfather pendulum as a simple pendulum really.

As a historical aside, James Cox of London built a self-winding grandfather clock that was powered by harvesting the energy from daily changes in air pressure using the rising and falling of a mercury barometer. This was done in the 1760's.

https://en.wikipedia.org/wiki/Cox's_timepiece

 

[sophiecentaur]

I searched this thread for the term "restoring force" and couldn't find it. It's worth pointing out that True Simple Harmonic Motion only occurs when the restoring force is proportional the negative of the displacement. There's no more to say, really because the restoring force for a simple pendulum is not linear with respect to displacement.

 

[sophiecentaur]

Grandfather clocks also have a mechanism for "brining back" energy into the system (anchor escapement).

Isn't that true for any mechanical clock mechanism? If you look at the wheels on an ordinary pendulum clock, they don't just move forward in jerks but they move backwards a bit before moving forwards. It's too fast to see on my tick tick watch escapement.
I wouldn't mind betting it boils down to impedance matching the source to the losses in the mechanism - same as with a high stability electrical oscillator.

 

[malawi_glenn]

Isn't that true for any mechanical clock mechanism?

I have no idea. What is an "ordinary pendulum clock" btw?

 

[Baluncore]

I wouldn't mind betting it boils down to impedance matching the source to the losses in the mechanism - same as with a high stability electrical oscillator.

The pendulum amplitude must be regulated to maintain a regular natural frequency. That is usually done by the dimensions of the escapement, but then the energy balance is made up at the extremes of the pendulum swing, when the velocity is near zero. In order to not influence the period, the energy losses must be made-up at the zero crossings of the pendulum. Regulation of the amplitude by an escapement, while replacement of the lost energy by the same escapement, are therefore at odds, which makes every mechanical pendulum clock a compromise. For a pendulum to have a regular period, the amplitude can be sensed optically, and the missing energy replaced electromagnetically, one quarter of a cycle later, as the pendulum passes vertical. When compared to GPS time, the rate of the pendulum should then show variation due to the lunar and solar tides.

 

[anorlunda]

Clock escapements existed long before electromagnets.

In the text below, I bolded the portion that deals with energy replacement. The replacement occurs near the bottom of the swing, not the extremes.

https://en.wikipedia.org/wiki/Anchor_escapement​

The deadbeat escapement has two faces to the pallets, a 'locking' or 'dead' face, with a curved surface concentric with the axis on which the anchor rotates, and a sloping 'impulse' face.[3] When an escape wheel tooth is resting against one of the dead faces, its force is directed through the anchor's pivot axis, so it gives no impulse to the pendulum, allowing it to swing freely. When the pallet on the other side releases the escape wheel, a tooth lands on this "dead" face first, and remains resting against it for most of the pendulum's outward swing and return. For this period the escape wheel is "locked" and unable to turn. Near the bottom of the pendulum's swing the tooth slides off the dead face onto the slanted 'impulse' face of the pallet, allowing the escape wheel to turn and give the pendulum a push, before dropping off the pallet. It is still a frictional rest escapement because the sliding of the escape tooth on the dead face adds friction to the pendulum's swing, but it has less friction than the recoil escapement because there is no recoil force.

 

[vanhees71]

I searched this thread for the term "restoring force" and couldn't find it. It's worth pointing out that True Simple Harmonic Motion only occurs when the restoring force is proportional the negative of the displacement. There's no more to say, really because the restoring force for a simple pendulum is not linear with respect to displacement.

I used "driving force" instead ;-))

 

[malawi_glenn]

There's no more to say, really because the restoring force for a simple pendulum is not linear with respect to displacement.

But in ALL intro college physics books, the simple pendulum is treated like an approximate HO due to the small angle approximation.

I'd say that this is also true for physical springs too, the restoring force is only linear up to a certain point.

It's just mathematical models of physical systems anyway.

Same with QM HO too, approximating the force between atoms in say a diatomic molecule, or between atoms in a solid material lattice. For small excitation energies, the HO is a good enough approximation. But it is still an approximation, a very useful though.

 

[PeroK]

The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.

Sidney Coleman

 

[malawi_glenn]

The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.

Sidney Coleman

Sidney Coleman is the most underappriciated physicist ever.
Reminder to myself: need to get his collection of QFT and relativity lectures...

 

[vanhees71]

I just try to get his book on QFT for quite a while now. The first attempt to order it from amazon and then from another German online bookseller failed. Then I tried it directly from the publisher (World Scientific), which also was no success. Now I ordered it again from the online bookseller. They say it'll take "more than 4 weeks" to order it. We'll see... It's a shame that this brillant book is not available in print (I've the e-book to know that it is among the best intro QFT books I've seen yet).

 

[vanhees71]

The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.

Sidney Coleman

That's true. Almost all analytically solvable problems in classical as well as quantum physics are in some way related to the harmonic oscillator ;-)).

 

[malawi_glenn]

I just try to get his book on QFT for quite a while now. The first attempt to order it from amazon and then from another German online bookseller failed. Then I tried it directly from the publisher (World Scientific)

Ah that sounds bad.
I should keep this in mind, I had some issues with Springer during corona with printing. They said they were out of paper! Someone at my school said it was probably related to some strike in finland which is a very big paper manufacturer country.
Aspects of symmetry is on of my top 3 or 5 physics books.

Hope you will be able to get his QFT book soon!

 

[sophiecentaur]

I used "driving force" instead ;-))

Hmm. If I were asked to point at the "driving force" of a pendulum, I would say it's the weight force, pulling the bob downwards but the term 'restoring force' is firm embedded in my memory so I may have over-reacted.

 

[malawi_glenn]

I often picture "driving force" as some kind of motor or similar that is feeding energy into the system

 

[sophiecentaur]

I have no idea. What is an "ordinary pendulum clock" btw?

I used the word 'ordinary' because, afaik, the Grandfather (or Long Case Clock) is not particularly special; it just has a longer pendulum than usual and, I believe they are all powered by falling weights.
The 'energy harvesting' clock that you quote is interesting- more of a novelty perhaps than a strict time keeper. If you are relying on temperature variation to provide Power then I suspect that could be a built-in design feature that could affect timekeeping accuracy. The system of falling weights is good in that the driving force is much more constant than a simple (Hooke's Law) spring. There are some very clever methods of regulating it.

 

[anorlunda]

If you are relying on temperature variation to provide Power then I suspect that could be a built-in design feature that could affect timekeeping accuracy. The system of falling weights is good in that the driving force is much more constant than a simple (Hooke's Law) spring.

Atmospheric clocks use harvested energy to rewind the spring, not to drive the clock directly.

Also, the goal of the art and science of clockmaking was for centuries devoted to divorcing variations in driving force, from the speed of the clock. Foremost among variable forces is gravity, because clocks and watches were required to be accurate even with variable orientations. Even stage coach clocks had to resist the forces of bumpy roads. (See the book Longitude.)

My father was a clockmaker. He would roll over in his grave if he heard you say that constant driving force was important.

The whole idea of the escapement mechanism (see #17) is to control the speed independent of the driving force. It was invented in the 13th century for that exact purpose.

 

[sophiecentaur]

You are pretty much right but I have a so-called school clock which is very basic but its most expensive component is the fusee system which is a ‘curved’ conical drum with a fine chain wrapped round it.
As the spring winds down, the radius reduces to keep the torque constant [Edit: or it may be the other way round, depending on what drives what]. The clock maintains its time to within one minute a week but then it runs fast.
No doubt the clocks you are describing are inherently better, even without the fusee.
If it were not such a fiddly business, I’d like to ‘do’ clocks. (Proper ones, I mean)

 

[malawi_glenn]

We need a "pendulum clock" thread and insight article :)