How does the accuracy of the clock change when the spring stretches?

[tomceka]

Homework Statement

A pendulum clock - is a clock that uses a pendulum, a swinging weight, as its timekeeping element.

Pendulum's amplitude of oscillation decreases with the stretching of the spring. How does the accuracy of the clock change?

Relevant Equations

1) T=2π√(m/k), T - period of oscillation, k - spring constant
2) T=2π√(l/g),

A longer pendulum swings slower. So changing the length l of the pendulum changes the period T, which affects the timekeeping accuracy. But the problem is talking about the body on the spring, not the string. So the second formula cannot be applied here directly and I don't know how to progress further since the first equation only involves mass and the coefficient which we do not know.

 

[kuruman]

Where is the spring in this picture? Is the bob attached to a pendulum through a spring? I doubt it. Can you post the entire question, specifically the part where "the problem is talking about the body on the spring"?

 

[hutchphd]

On an escapement drive typically the pendulum amplitude is slightly dependent on spring tension hence

"Pendulum's amplitude of oscillation decreases with the stretching of the spring. How does the accuracy of the clock change?"

 

[haruspex]

On an escapement drive typically the pendulum amplitude is slightly dependent on spring tension hence

"Pendulum's amplitude of oscillation decreases with the stretching of the spring. How does the accuracy of the clock change?"

Yes, it's clearly concerned with the amplitude variation.

@tomceka , you need to go back to the exact form of the pendulum motion equation, the form that still has $\sin(\theta)$ in it. The standard solution, turning it into the SHM equation, results from approximating it as $\theta$. Suppose we were to keep the next term of the power expansion, $-\frac 16\theta^3$. Can you figure out roughly how changing the amplitude would affect the period? I think it's not easy.

(Edit: see http://hyperphysics.phy-astr.gsu.edu/hbase/pendl.html)

That said, I am still puzzled by the reference to "stretching" of the spring. Assuming a standard coiled clockwork spring, the spring unwinds and relaxes as it runs down; it never stretches.

 

[hutchphd]

I think it's not easy.

That said, I am still puzzled by the reference to "stretching" of the spring. Assuming a standard coiled clockwork spring, the spring unwinds and relaxes as it runs down; it never stretches.

I agree with both observations. Seems an odd question with little guidance (for an introductory course).

 

[erobz]

Is it possible that instead of a fixed rod on the pendulum, it is instead a mass on a spring swinging…that would be more advanced. Or, maybe they introduce the spring as a mechanism to “change the length” of the rod in the problem instead of saying the rod just increases in length, decreasing the amplitude.

 

[Delta2]

On an escapement drive typically

Ah it is the well known escapement drive aka warp drive

 

[haruspex]

Is it possible that instead of a fixed rod on the pendulum, it is instead a mass on a spring swinging…that would be more advanced. Or, maybe they introduce the spring as a mechanism to “change the length” of the rod in the problem instead of saying the rod just increases in length, decreasing the amplitude.

The question states that some change in a spring affects amplitude, and, as I read it, then asks how an amplitude change may affect the period. If so, the correctness of the first part is somewhat academic; we only have to address the second.

 

[nasu]

Some pendulum clocks use a spring to store energy. Winding the clock in this case means that a spring is distorted such that it stores energy. Usually a spiral spring, so it is not really "stretching". The spring acts the mechanism that drives the pendulum. The spring has nothing to do with the period, at least not directly. But as the force of the spring may decrease towards the end of the winding cycle it may also change a little the exact positions in which the pendulum is acted/released.

 

[haruspex]

as the force of the spring may decrease towards the end of the winding cycle it may also change a little the exact positions in which the pendulum is acted/released.

It will reduce the amplitude since the torque on the pendulum is weakening. This aspect seems to be the one referred to in the question.

 

[nasu]

The question seem to be how is the accuracy affected by a change in amplitude. Not how this change in amplitude happens exactly. It would help if they define "accuarcy" for this case.

 

[haruspex]

The question seem to be how is the accuracy affected by a change in amplitude. Not how this change in amplitude happens exactly.

… as I wrote in post #8.

It would help if they define "accuarcy" for this case.

Good point.. it would be better to ask how it affects the rate.

 

[Delta2]

Is this about a pendulum who uses some sort of spring mechanism to recover the energy lost due to friction? So that the pendulum keeps swinging?

 

[haruspex]

Is this about a pendulum who uses some sort of spring mechanism to recover the energy lost due to friction? So that the pendulum keeps swinging?

https://www.scienceabc.com/innovati...-escapement-ticking-sound-falling-weight.html

 

[nasu]

… as I wrote in post #8.

Good point.. it would be better to ask how it affects the rate.

Affecting the accuracy isn't necesarilly the same as affecting the period of the pendulum (mentioned in post 8). Maybe it is. This is why I said that we need to know what they mean by this.

 

[hutchphd]

Affecting the accuracy isn't necesarilly the same as affecting the period of the pendulum (mentioned in post 8). Maybe it is. This is why I said that we need to know what they mean by this.

If your clock is not broken, yes it is. That's why the clock has a pendulum.

 

[nasu]

If the escapment mechanism is released every two periods instead of one the accuracy is affected even though the period is not. It does not mean that the clock is broken But I don't think this is possible. Nevermind.

 

[hutchphd]

And by some measures a stopped clock is more accurate than one that loses a millisecond a day. But really...

 

[Delta2]

Have you heard of classical microcombinative theory?

 

[haruspex]

If your clock is not broken, yes it is. That's why the clock has a pendulum.

The reason I agreed with @nasu on this point in post #12 was that the clock might have been running too fast initially, so as the rate slows the accuracy improves. I have such a clock; I deliberately have it gaining slightly when fully wound so that it only loses slightly later on.

 

[Delta2]

I have such a clock;

Do you have it from the 70-80s era?

 

[kuruman]

I have such a clock; I deliberately have it gaining slightly when fully wound so that it only loses slightly later on.

So your clock tells the time accurately only once between windings. Have you considered stopping it? Then you're guaranteed accurate readings twice a day and you don't have to wind it.

 

[haruspex]

So your clock tells the time accurately only once between windings. Have you considered stopping it? Then you're guaranteed accurate readings twice a day and you don't have to wind it.

When I wind it, it is slightly behind, so then catches up and gets ahead. As the spring unwinds, it slows down and gets behind again. So it‘s right twice per winding… but that's only once a week. It's German, 1920s or earlier.
My Welsh longcase clock, ca. 1750, doesn't have that problem, the motive force being constant. Accurate to within a minute over a week!

 

[Delta2]

So your clock tells the time accurately only once between windings. Have you considered stopping it? Then you're guaranteed accurate readings twice a day and you don't have to wind it.

what if he stops it right at the time its reading becomes accurate?